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			650 lines
		
	
	
		
			18 KiB
		
	
	
	
		
			Go
		
	
	
	
	
	
			
		
		
	
	
			650 lines
		
	
	
		
			18 KiB
		
	
	
	
		
			Go
		
	
	
	
	
	
| // Copyright 2014 Google Inc.
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| //
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| // Licensed under the Apache License, Version 2.0 (the "License");
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| // you may not use this file except in compliance with the License.
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| // You may obtain a copy of the License at
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| //
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| //     http://www.apache.org/licenses/LICENSE-2.0
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| //
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| // Unless required by applicable law or agreed to in writing, software
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| // distributed under the License is distributed on an "AS IS" BASIS,
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| // WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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| // See the License for the specific language governing permissions and
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| // limitations under the License.
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| 
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| // Package btree implements in-memory B-Trees of arbitrary degree.
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| //
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| // btree implements an in-memory B-Tree for use as an ordered data structure.
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| // It is not meant for persistent storage solutions.
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| //
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| // It has a flatter structure than an equivalent red-black or other binary tree,
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| // which in some cases yields better memory usage and/or performance.
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| // See some discussion on the matter here:
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| //   http://google-opensource.blogspot.com/2013/01/c-containers-that-save-memory-and-time.html
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| // Note, though, that this project is in no way related to the C++ B-Tree
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| // implmentation written about there.
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| //
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| // Within this tree, each node contains a slice of items and a (possibly nil)
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| // slice of children.  For basic numeric values or raw structs, this can cause
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| // efficiency differences when compared to equivalent C++ template code that
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| // stores values in arrays within the node:
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| //   * Due to the overhead of storing values as interfaces (each
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| //     value needs to be stored as the value itself, then 2 words for the
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| //     interface pointing to that value and its type), resulting in higher
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| //     memory use.
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| //   * Since interfaces can point to values anywhere in memory, values are
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| //     most likely not stored in contiguous blocks, resulting in a higher
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| //     number of cache misses.
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| // These issues don't tend to matter, though, when working with strings or other
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| // heap-allocated structures, since C++-equivalent structures also must store
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| // pointers and also distribute their values across the heap.
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| //
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| // This implementation is designed to be a drop-in replacement to gollrb.LLRB
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| // trees, (http://github.com/petar/gollrb), an excellent and probably the most
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| // widely used ordered tree implementation in the Go ecosystem currently.
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| // Its functions, therefore, exactly mirror those of
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| // llrb.LLRB where possible.  Unlike gollrb, though, we currently don't
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| // support storing multiple equivalent values or backwards iteration.
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| package btree
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| 
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| import (
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| 	"fmt"
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| 	"io"
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| 	"sort"
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| 	"strings"
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| )
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| 
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| // Item represents a single object in the tree.
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| type Item interface {
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| 	// Less tests whether the current item is less than the given argument.
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| 	//
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| 	// This must provide a strict weak ordering.
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| 	// If !a.Less(b) && !b.Less(a), we treat this to mean a == b (i.e. we can only
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| 	// hold one of either a or b in the tree).
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| 	Less(than Item) bool
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| }
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| 
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| const (
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| 	DefaultFreeListSize = 32
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| )
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| 
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| // FreeList represents a free list of btree nodes. By default each
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| // BTree has its own FreeList, but multiple BTrees can share the same
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| // FreeList.
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| // Two Btrees using the same freelist are not safe for concurrent write access.
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| type FreeList struct {
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| 	freelist []*node
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| }
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| 
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| // NewFreeList creates a new free list.
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| // size is the maximum size of the returned free list.
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| func NewFreeList(size int) *FreeList {
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| 	return &FreeList{freelist: make([]*node, 0, size)}
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| }
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| 
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| func (f *FreeList) newNode() (n *node) {
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| 	index := len(f.freelist) - 1
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| 	if index < 0 {
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| 		return new(node)
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| 	}
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| 	f.freelist, n = f.freelist[:index], f.freelist[index]
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| 	return
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| }
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| 
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| func (f *FreeList) freeNode(n *node) {
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| 	if len(f.freelist) < cap(f.freelist) {
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| 		f.freelist = append(f.freelist, n)
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| 	}
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| }
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| 
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| // ItemIterator allows callers of Ascend* to iterate in-order over portions of
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| // the tree.  When this function returns false, iteration will stop and the
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| // associated Ascend* function will immediately return.
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| type ItemIterator func(i Item) bool
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| 
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| // New creates a new B-Tree with the given degree.
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| //
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| // New(2), for example, will create a 2-3-4 tree (each node contains 1-3 items
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| // and 2-4 children).
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| func New(degree int) *BTree {
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| 	return NewWithFreeList(degree, NewFreeList(DefaultFreeListSize))
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| }
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| 
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| // NewWithFreeList creates a new B-Tree that uses the given node free list.
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| func NewWithFreeList(degree int, f *FreeList) *BTree {
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| 	if degree <= 1 {
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| 		panic("bad degree")
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| 	}
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| 	return &BTree{
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| 		degree:   degree,
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| 		freelist: f,
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| 	}
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| }
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| 
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| // items stores items in a node.
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| type items []Item
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| 
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| // insertAt inserts a value into the given index, pushing all subsequent values
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| // forward.
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| func (s *items) insertAt(index int, item Item) {
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| 	*s = append(*s, nil)
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| 	if index < len(*s) {
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| 		copy((*s)[index+1:], (*s)[index:])
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| 	}
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| 	(*s)[index] = item
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| }
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| 
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| // removeAt removes a value at a given index, pulling all subsequent values
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| // back.
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| func (s *items) removeAt(index int) Item {
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| 	item := (*s)[index]
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| 	(*s)[index] = nil
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| 	copy((*s)[index:], (*s)[index+1:])
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| 	*s = (*s)[:len(*s)-1]
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| 	return item
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| }
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| 
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| // pop removes and returns the last element in the list.
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| func (s *items) pop() (out Item) {
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| 	index := len(*s) - 1
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| 	out = (*s)[index]
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| 	(*s)[index] = nil
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| 	*s = (*s)[:index]
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| 	return
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| }
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| 
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| // find returns the index where the given item should be inserted into this
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| // list.  'found' is true if the item already exists in the list at the given
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| // index.
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| func (s items) find(item Item) (index int, found bool) {
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| 	i := sort.Search(len(s), func(i int) bool {
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| 		return item.Less(s[i])
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| 	})
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| 	if i > 0 && !s[i-1].Less(item) {
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| 		return i - 1, true
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| 	}
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| 	return i, false
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| }
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| 
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| // children stores child nodes in a node.
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| type children []*node
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| 
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| // insertAt inserts a value into the given index, pushing all subsequent values
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| // forward.
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| func (s *children) insertAt(index int, n *node) {
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| 	*s = append(*s, nil)
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| 	if index < len(*s) {
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| 		copy((*s)[index+1:], (*s)[index:])
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| 	}
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| 	(*s)[index] = n
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| }
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| 
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| // removeAt removes a value at a given index, pulling all subsequent values
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| // back.
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| func (s *children) removeAt(index int) *node {
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| 	n := (*s)[index]
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| 	(*s)[index] = nil
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| 	copy((*s)[index:], (*s)[index+1:])
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| 	*s = (*s)[:len(*s)-1]
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| 	return n
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| }
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| 
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| // pop removes and returns the last element in the list.
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| func (s *children) pop() (out *node) {
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| 	index := len(*s) - 1
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| 	out = (*s)[index]
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| 	(*s)[index] = nil
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| 	*s = (*s)[:index]
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| 	return
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| }
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| 
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| // node is an internal node in a tree.
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| //
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| // It must at all times maintain the invariant that either
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| //   * len(children) == 0, len(items) unconstrained
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| //   * len(children) == len(items) + 1
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| type node struct {
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| 	items    items
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| 	children children
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| 	t        *BTree
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| }
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| 
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| // split splits the given node at the given index.  The current node shrinks,
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| // and this function returns the item that existed at that index and a new node
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| // containing all items/children after it.
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| func (n *node) split(i int) (Item, *node) {
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| 	item := n.items[i]
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| 	next := n.t.newNode()
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| 	next.items = append(next.items, n.items[i+1:]...)
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| 	n.items = n.items[:i]
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| 	if len(n.children) > 0 {
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| 		next.children = append(next.children, n.children[i+1:]...)
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| 		n.children = n.children[:i+1]
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| 	}
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| 	return item, next
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| }
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| 
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| // maybeSplitChild checks if a child should be split, and if so splits it.
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| // Returns whether or not a split occurred.
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| func (n *node) maybeSplitChild(i, maxItems int) bool {
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| 	if len(n.children[i].items) < maxItems {
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| 		return false
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| 	}
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| 	first := n.children[i]
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| 	item, second := first.split(maxItems / 2)
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| 	n.items.insertAt(i, item)
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| 	n.children.insertAt(i+1, second)
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| 	return true
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| }
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| 
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| // insert inserts an item into the subtree rooted at this node, making sure
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| // no nodes in the subtree exceed maxItems items.  Should an equivalent item be
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| // be found/replaced by insert, it will be returned.
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| func (n *node) insert(item Item, maxItems int) Item {
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| 	i, found := n.items.find(item)
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| 	if found {
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| 		out := n.items[i]
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| 		n.items[i] = item
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| 		return out
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| 	}
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| 	if len(n.children) == 0 {
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| 		n.items.insertAt(i, item)
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| 		return nil
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| 	}
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| 	if n.maybeSplitChild(i, maxItems) {
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| 		inTree := n.items[i]
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| 		switch {
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| 		case item.Less(inTree):
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| 			// no change, we want first split node
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| 		case inTree.Less(item):
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| 			i++ // we want second split node
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| 		default:
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| 			out := n.items[i]
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| 			n.items[i] = item
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| 			return out
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| 		}
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| 	}
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| 	return n.children[i].insert(item, maxItems)
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| }
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| 
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| // get finds the given key in the subtree and returns it.
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| func (n *node) get(key Item) Item {
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| 	i, found := n.items.find(key)
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| 	if found {
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| 		return n.items[i]
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| 	} else if len(n.children) > 0 {
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| 		return n.children[i].get(key)
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| 	}
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| 	return nil
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| }
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| 
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| // min returns the first item in the subtree.
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| func min(n *node) Item {
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| 	if n == nil {
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| 		return nil
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| 	}
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| 	for len(n.children) > 0 {
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| 		n = n.children[0]
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| 	}
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| 	if len(n.items) == 0 {
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| 		return nil
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| 	}
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| 	return n.items[0]
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| }
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| 
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| // max returns the last item in the subtree.
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| func max(n *node) Item {
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| 	if n == nil {
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| 		return nil
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| 	}
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| 	for len(n.children) > 0 {
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| 		n = n.children[len(n.children)-1]
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| 	}
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| 	if len(n.items) == 0 {
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| 		return nil
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| 	}
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| 	return n.items[len(n.items)-1]
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| }
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| 
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| // toRemove details what item to remove in a node.remove call.
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| type toRemove int
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| 
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| const (
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| 	removeItem toRemove = iota // removes the given item
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| 	removeMin                  // removes smallest item in the subtree
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| 	removeMax                  // removes largest item in the subtree
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| )
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| 
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| // remove removes an item from the subtree rooted at this node.
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| func (n *node) remove(item Item, minItems int, typ toRemove) Item {
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| 	var i int
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| 	var found bool
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| 	switch typ {
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| 	case removeMax:
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| 		if len(n.children) == 0 {
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| 			return n.items.pop()
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| 		}
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| 		i = len(n.items)
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| 	case removeMin:
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| 		if len(n.children) == 0 {
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| 			return n.items.removeAt(0)
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| 		}
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| 		i = 0
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| 	case removeItem:
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| 		i, found = n.items.find(item)
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| 		if len(n.children) == 0 {
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| 			if found {
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| 				return n.items.removeAt(i)
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| 			}
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| 			return nil
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| 		}
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| 	default:
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| 		panic("invalid type")
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| 	}
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| 	// If we get to here, we have children.
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| 	child := n.children[i]
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| 	if len(child.items) <= minItems {
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| 		return n.growChildAndRemove(i, item, minItems, typ)
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| 	}
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| 	// Either we had enough items to begin with, or we've done some
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| 	// merging/stealing, because we've got enough now and we're ready to return
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| 	// stuff.
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| 	if found {
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| 		// The item exists at index 'i', and the child we've selected can give us a
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| 		// predecessor, since if we've gotten here it's got > minItems items in it.
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| 		out := n.items[i]
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| 		// We use our special-case 'remove' call with typ=maxItem to pull the
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| 		// predecessor of item i (the rightmost leaf of our immediate left child)
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| 		// and set it into where we pulled the item from.
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| 		n.items[i] = child.remove(nil, minItems, removeMax)
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| 		return out
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| 	}
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| 	// Final recursive call.  Once we're here, we know that the item isn't in this
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| 	// node and that the child is big enough to remove from.
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| 	return child.remove(item, minItems, typ)
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| }
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| 
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| // growChildAndRemove grows child 'i' to make sure it's possible to remove an
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| // item from it while keeping it at minItems, then calls remove to actually
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| // remove it.
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| //
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| // Most documentation says we have to do two sets of special casing:
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| //   1) item is in this node
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| //   2) item is in child
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| // In both cases, we need to handle the two subcases:
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| //   A) node has enough values that it can spare one
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| //   B) node doesn't have enough values
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| // For the latter, we have to check:
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| //   a) left sibling has node to spare
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| //   b) right sibling has node to spare
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| //   c) we must merge
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| // To simplify our code here, we handle cases #1 and #2 the same:
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| // If a node doesn't have enough items, we make sure it does (using a,b,c).
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| // We then simply redo our remove call, and the second time (regardless of
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| // whether we're in case 1 or 2), we'll have enough items and can guarantee
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| // that we hit case A.
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| func (n *node) growChildAndRemove(i int, item Item, minItems int, typ toRemove) Item {
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| 	child := n.children[i]
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| 	if i > 0 && len(n.children[i-1].items) > minItems {
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| 		// Steal from left child
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| 		stealFrom := n.children[i-1]
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| 		stolenItem := stealFrom.items.pop()
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| 		child.items.insertAt(0, n.items[i-1])
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| 		n.items[i-1] = stolenItem
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| 		if len(stealFrom.children) > 0 {
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| 			child.children.insertAt(0, stealFrom.children.pop())
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| 		}
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| 	} else if i < len(n.items) && len(n.children[i+1].items) > minItems {
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| 		// steal from right child
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| 		stealFrom := n.children[i+1]
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| 		stolenItem := stealFrom.items.removeAt(0)
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| 		child.items = append(child.items, n.items[i])
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| 		n.items[i] = stolenItem
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| 		if len(stealFrom.children) > 0 {
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| 			child.children = append(child.children, stealFrom.children.removeAt(0))
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| 		}
 | |
| 	} else {
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| 		if i >= len(n.items) {
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| 			i--
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| 			child = n.children[i]
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| 		}
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| 		// merge with right child
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| 		mergeItem := n.items.removeAt(i)
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| 		mergeChild := n.children.removeAt(i + 1)
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| 		child.items = append(child.items, mergeItem)
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| 		child.items = append(child.items, mergeChild.items...)
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| 		child.children = append(child.children, mergeChild.children...)
 | |
| 		n.t.freeNode(mergeChild)
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| 	}
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| 	return n.remove(item, minItems, typ)
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| }
 | |
| 
 | |
| // iterate provides a simple method for iterating over elements in the tree.
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| // It could probably use some work to be extra-efficient (it calls from() a
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| // little more than it should), but it works pretty well for now.
 | |
| //
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| // It requires that 'from' and 'to' both return true for values we should hit
 | |
| // with the iterator.  It should also be the case that 'from' returns true for
 | |
| // values less than or equal to values 'to' returns true for, and 'to'
 | |
| // returns true for values greater than or equal to those that 'from'
 | |
| // does.
 | |
| func (n *node) iterate(from, to func(Item) bool, iter ItemIterator) bool {
 | |
| 	for i, item := range n.items {
 | |
| 		if !from(item) {
 | |
| 			continue
 | |
| 		}
 | |
| 		if len(n.children) > 0 && !n.children[i].iterate(from, to, iter) {
 | |
| 			return false
 | |
| 		}
 | |
| 		if !to(item) {
 | |
| 			return false
 | |
| 		}
 | |
| 		if !iter(item) {
 | |
| 			return false
 | |
| 		}
 | |
| 	}
 | |
| 	if len(n.children) > 0 {
 | |
| 		return n.children[len(n.children)-1].iterate(from, to, iter)
 | |
| 	}
 | |
| 	return true
 | |
| }
 | |
| 
 | |
| // Used for testing/debugging purposes.
 | |
| func (n *node) print(w io.Writer, level int) {
 | |
| 	fmt.Fprintf(w, "%sNODE:%v\n", strings.Repeat("  ", level), n.items)
 | |
| 	for _, c := range n.children {
 | |
| 		c.print(w, level+1)
 | |
| 	}
 | |
| }
 | |
| 
 | |
| // BTree is an implementation of a B-Tree.
 | |
| //
 | |
| // BTree stores Item instances in an ordered structure, allowing easy insertion,
 | |
| // removal, and iteration.
 | |
| //
 | |
| // Write operations are not safe for concurrent mutation by multiple
 | |
| // goroutines, but Read operations are.
 | |
| type BTree struct {
 | |
| 	degree   int
 | |
| 	length   int
 | |
| 	root     *node
 | |
| 	freelist *FreeList
 | |
| }
 | |
| 
 | |
| // maxItems returns the max number of items to allow per node.
 | |
| func (t *BTree) maxItems() int {
 | |
| 	return t.degree*2 - 1
 | |
| }
 | |
| 
 | |
| // minItems returns the min number of items to allow per node (ignored for the
 | |
| // root node).
 | |
| func (t *BTree) minItems() int {
 | |
| 	return t.degree - 1
 | |
| }
 | |
| 
 | |
| func (t *BTree) newNode() (n *node) {
 | |
| 	n = t.freelist.newNode()
 | |
| 	n.t = t
 | |
| 	return
 | |
| }
 | |
| 
 | |
| func (t *BTree) freeNode(n *node) {
 | |
| 	for i := range n.items {
 | |
| 		n.items[i] = nil // clear to allow GC
 | |
| 	}
 | |
| 	n.items = n.items[:0]
 | |
| 	for i := range n.children {
 | |
| 		n.children[i] = nil // clear to allow GC
 | |
| 	}
 | |
| 	n.children = n.children[:0]
 | |
| 	n.t = nil // clear to allow GC
 | |
| 	t.freelist.freeNode(n)
 | |
| }
 | |
| 
 | |
| // ReplaceOrInsert adds the given item to the tree.  If an item in the tree
 | |
| // already equals the given one, it is removed from the tree and returned.
 | |
| // Otherwise, nil is returned.
 | |
| //
 | |
| // nil cannot be added to the tree (will panic).
 | |
| func (t *BTree) ReplaceOrInsert(item Item) Item {
 | |
| 	if item == nil {
 | |
| 		panic("nil item being added to BTree")
 | |
| 	}
 | |
| 	if t.root == nil {
 | |
| 		t.root = t.newNode()
 | |
| 		t.root.items = append(t.root.items, item)
 | |
| 		t.length++
 | |
| 		return nil
 | |
| 	} else if len(t.root.items) >= t.maxItems() {
 | |
| 		item2, second := t.root.split(t.maxItems() / 2)
 | |
| 		oldroot := t.root
 | |
| 		t.root = t.newNode()
 | |
| 		t.root.items = append(t.root.items, item2)
 | |
| 		t.root.children = append(t.root.children, oldroot, second)
 | |
| 	}
 | |
| 	out := t.root.insert(item, t.maxItems())
 | |
| 	if out == nil {
 | |
| 		t.length++
 | |
| 	}
 | |
| 	return out
 | |
| }
 | |
| 
 | |
| // Delete removes an item equal to the passed in item from the tree, returning
 | |
| // it.  If no such item exists, returns nil.
 | |
| func (t *BTree) Delete(item Item) Item {
 | |
| 	return t.deleteItem(item, removeItem)
 | |
| }
 | |
| 
 | |
| // DeleteMin removes the smallest item in the tree and returns it.
 | |
| // If no such item exists, returns nil.
 | |
| func (t *BTree) DeleteMin() Item {
 | |
| 	return t.deleteItem(nil, removeMin)
 | |
| }
 | |
| 
 | |
| // DeleteMax removes the largest item in the tree and returns it.
 | |
| // If no such item exists, returns nil.
 | |
| func (t *BTree) DeleteMax() Item {
 | |
| 	return t.deleteItem(nil, removeMax)
 | |
| }
 | |
| 
 | |
| func (t *BTree) deleteItem(item Item, typ toRemove) Item {
 | |
| 	if t.root == nil || len(t.root.items) == 0 {
 | |
| 		return nil
 | |
| 	}
 | |
| 	out := t.root.remove(item, t.minItems(), typ)
 | |
| 	if len(t.root.items) == 0 && len(t.root.children) > 0 {
 | |
| 		oldroot := t.root
 | |
| 		t.root = t.root.children[0]
 | |
| 		t.freeNode(oldroot)
 | |
| 	}
 | |
| 	if out != nil {
 | |
| 		t.length--
 | |
| 	}
 | |
| 	return out
 | |
| }
 | |
| 
 | |
| // AscendRange calls the iterator for every value in the tree within the range
 | |
| // [greaterOrEqual, lessThan), until iterator returns false.
 | |
| func (t *BTree) AscendRange(greaterOrEqual, lessThan Item, iterator ItemIterator) {
 | |
| 	if t.root == nil {
 | |
| 		return
 | |
| 	}
 | |
| 	t.root.iterate(
 | |
| 		func(a Item) bool { return !a.Less(greaterOrEqual) },
 | |
| 		func(a Item) bool { return a.Less(lessThan) },
 | |
| 		iterator)
 | |
| }
 | |
| 
 | |
| // AscendLessThan calls the iterator for every value in the tree within the range
 | |
| // [first, pivot), until iterator returns false.
 | |
| func (t *BTree) AscendLessThan(pivot Item, iterator ItemIterator) {
 | |
| 	if t.root == nil {
 | |
| 		return
 | |
| 	}
 | |
| 	t.root.iterate(
 | |
| 		func(a Item) bool { return true },
 | |
| 		func(a Item) bool { return a.Less(pivot) },
 | |
| 		iterator)
 | |
| }
 | |
| 
 | |
| // AscendGreaterOrEqual calls the iterator for every value in the tree within
 | |
| // the range [pivot, last], until iterator returns false.
 | |
| func (t *BTree) AscendGreaterOrEqual(pivot Item, iterator ItemIterator) {
 | |
| 	if t.root == nil {
 | |
| 		return
 | |
| 	}
 | |
| 	t.root.iterate(
 | |
| 		func(a Item) bool { return !a.Less(pivot) },
 | |
| 		func(a Item) bool { return true },
 | |
| 		iterator)
 | |
| }
 | |
| 
 | |
| // Ascend calls the iterator for every value in the tree within the range
 | |
| // [first, last], until iterator returns false.
 | |
| func (t *BTree) Ascend(iterator ItemIterator) {
 | |
| 	if t.root == nil {
 | |
| 		return
 | |
| 	}
 | |
| 	t.root.iterate(
 | |
| 		func(a Item) bool { return true },
 | |
| 		func(a Item) bool { return true },
 | |
| 		iterator)
 | |
| }
 | |
| 
 | |
| // Get looks for the key item in the tree, returning it.  It returns nil if
 | |
| // unable to find that item.
 | |
| func (t *BTree) Get(key Item) Item {
 | |
| 	if t.root == nil {
 | |
| 		return nil
 | |
| 	}
 | |
| 	return t.root.get(key)
 | |
| }
 | |
| 
 | |
| // Min returns the smallest item in the tree, or nil if the tree is empty.
 | |
| func (t *BTree) Min() Item {
 | |
| 	return min(t.root)
 | |
| }
 | |
| 
 | |
| // Max returns the largest item in the tree, or nil if the tree is empty.
 | |
| func (t *BTree) Max() Item {
 | |
| 	return max(t.root)
 | |
| }
 | |
| 
 | |
| // Has returns true if the given key is in the tree.
 | |
| func (t *BTree) Has(key Item) bool {
 | |
| 	return t.Get(key) != nil
 | |
| }
 | |
| 
 | |
| // Len returns the number of items currently in the tree.
 | |
| func (t *BTree) Len() int {
 | |
| 	return t.length
 | |
| }
 | |
| 
 | |
| // Int implements the Item interface for integers.
 | |
| type Int int
 | |
| 
 | |
| // Less returns true if int(a) < int(b).
 | |
| func (a Int) Less(b Item) bool {
 | |
| 	return a < b.(Int)
 | |
| }
 |