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
}