二叉搜索树实现 in Go语言

用Go实现了下二叉搜索树，还是花了不少时间，在实现中使用的是单向链表，这才算是体会到了双向链表在实现中的优势

package datastructure

import (
"container/list"
"fmt"
)

type BSTree struct {
root *Node
}

type Node struct {
left *Node
right *Node
value int
}

// NewBSTree 创建树
func NewBSTree() *BSTree {
return &BSTree{}
}

func (t *BSTree) Insert(value int) {
var parent *Node
z := &Node{value: value}
x := t.root
for x != nil {
parent = x
if z.value < x.value {
x = x.left
} else {
x = x.right
}
}
if parent == nil { //该树为空
t.root = z
} else if z.value < parent.value {
parent.left = z
} else {
parent.right = z
}
}

func (t *BSTree) Search(x int) *Node {
node := t.root
for node != nil {
if node.value == x {
return node
} else if x < node.value {
node = node.left
} else {
node = node.right
}
}
return nil
}

func (t *BSTree) Delete(x int) bool {
var parent *Node
node := t.root
isFind := false
for node != nil {
if node.value == x {
isFind = true
break
} else if x < node.value {
parent = node
node = node.left
} else {
parent = node
node = node.right
}
}
if isFind == false {
return false
}
//情况一：node为叶节点
if node.left == nil && node.right == nil {
if parent == nil {
t.root = nil
} else {
if parent.left == node {
parent.left = nil
} else {
parent.right = nil
}
}
return true
}
//情况二:左孩子边为空或右边孩子为空
if node.left == nil || node.right == nil {
if parent == nil {
if node.left == nil {
t.root = node.right
} else {
t.root = node.left
}
} else {
if parent.left == node {
if node.left == nil {
parent.left = node.right
} else {
parent.left = node.left
}
} else {
if node.left == nil {
parent.right = node.right
} else {
parent.right = node.left
}
}
}
return true
}
//情况三：两个孩子都不为空
re := node.left
re_parent := node
for re.right != nil { //找到前驱节点和前驱节点的父节点
re_parent = re
re = re.right
}
node.value = re.value
if node == re_parent {
node.left = re.left
} else {
re_parent.right = re.left
}
return true
}

//PrintTree1 递归结构
func (t *Node) PrintTree1() {
if t.left != nil {
t.left.PrintTree1()
}
fmt.Print(t.value," ")
if t.right != nil {
t.right.PrintTree1()
}
}

//PrintTree2 非递归结构
func (t *Node) PrintTree2() {
stack := list.New()
stack.PushBack(t)
for {
node := stack.Back()
if node == nil {
return
}
stack.Remove(node)
v, _ := node.Value.(*Node)
fmt.Println(v.value)
if v.left != nil {
stack.PushBack(v.left)
}
if v.right != nil {
stack.PushBack(v.right)
}
}
}