福大大架构师每日一题

福大大架构师每日一题

2021-04-12:判断二叉树是否是搜索二叉树?

福大大 答案2021-04-12:

中序遍历有序即可。
1.递归。
2.莫里斯遍历。

代码用golang编写。代码如下:

package main

import "fmt"

const INT_MAX = int(^uint(0) >> 1)
const INT_MIN = ^INT_MAX

func main() {
   
    head := &TreeNode{
   Val: 5}
    head.Left = &TreeNode{
   Val: 3}
    head.Right = &TreeNode{
   Val: 7}
    head.Left.Left = &TreeNode{
   Val: 2}
    head.Left.Right = &TreeNode{
   Val: 4}
    head.Right.Left = &TreeNode{
   Val: 6}
    head.Right.Right = &TreeNode{
   Val: 8}
    ret := isBST1(head)
    fmt.Println("递归:", ret)
    fmt.Println("----")
    ret = isBST2(head)
    fmt.Println("莫里斯遍历:", ret)
}

//Definition for a binary tree node.
type TreeNode struct {
   
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func isBST1(head *TreeNode) bool {
   
    if head == nil {
   
        return true
    }
    ansVal := true
    ans := &ansVal
    preVal := INT_MIN
    pre := &preVal
    process(head, pre, ans)
    return *ans
}
func process(head *TreeNode, pre *int, ans *bool) {
   
    if head == nil {
   
        return
    }
    if *ans {
   
        process(head.Left, pre, ans)
    }
    if *ans {
   
        if *pre > head.Val {
   
            *ans = false
        } else {
   
            *pre = head.Val
        }
    }
    if *ans {
   
        process(head.Right, pre, ans)
    }
}

// 根据morris遍历改写
func isBST2(head *TreeNode) bool {
   
    if head == nil {
   
        return true
    }
    cur := head
    var mostRight *TreeNode
    preVal := INT_MIN
    for cur != nil {
   
        mostRight = cur.Left
        if mostRight != nil {
   
            for mostRight.Right != nil && mostRight.Right != cur {
   
                mostRight = mostRight.Right
            }
            if mostRight.Right == nil {
    //先 第一次到达
                mostRight.Right = cur
                cur = cur.Left
                continue
            } else {
    //后 第二次到达
                mostRight.Right = nil
            }
        } else {
    //先 只有一次到达
        }

        //此时cur就是中序
        if preVal > cur.Val {
   
            return false
        } else {
   
            preVal = cur.Val
        }

        //中
        cur = cur.Right

    }
    //后
    return true
}

执行结果如下:
2021-04-12:判断二叉树是否是搜索二叉树?-LMLPHP


左神java代码
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04-17 01:13