二叉搜索树实现 in Go语言

来源：互联网发布：数据库架构师工资编辑：程序博客网时间：2024/05/29 14:32

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 datastructureimport ("container/list""fmt")type BSTree struct {root *Node}type Node struct {left  *Noderight *Nodevalue int}// NewBSTree 创建树func NewBSTree() *BSTree {return &BSTree{}}func (t *BSTree) Insert(value int) {var parent *Nodez := &Node{value: value}x := t.rootfor x != nil {parent = xif 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.rootfor 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 *Nodenode := t.rootisFind := falsefor node != nil {if node.value == x {isFind = truebreak} else if x < node.value {parent = nodenode = node.left} else {parent = nodenode = 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)}}}

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