0098. Validate Binary Search Tree

# 98. Validate Binary Search Tree#

## 题目 #

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node’s key.
• The right subtree of a node contains only nodes with keys greater than the node’s key.
• Both the left and right subtrees must also be binary search trees.

xample 1:

``````    2
/ \
1   3

Input: [2,1,3]
Output: true
``````

Example 2:

``````    5
/ \
1   4
/ \
3   6

Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.
``````

## 题目大意 #

• 节点的左子树只包含小于当前节点的数。
• 节点的右子树只包含大于当前节点的数。
• 所有左子树和右子树自身必须也是二叉搜索树。

## 解题思路 #

• 判断一个树是否是 BST，按照定义递归判断即可

## 代码 #

``````
package leetcode

import "math"

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

// 解法一，直接按照定义比较大小，比 root 节点小的都在左边，比 root 节点大的都在右边
func isValidBST(root *TreeNode) bool {
return isValidbst(root, math.Inf(-1), math.Inf(1))
}
func isValidbst(root *TreeNode, min, max float64) bool {
if root == nil {
return true
}
v := float64(root.Val)
return v < max && v > min && isValidbst(root.Left, min, v) && isValidbst(root.Right, v, max)
}

// 解法二，把 BST 按照左中右的顺序输出到数组中，如果是 BST，则数组中的数字是从小到大有序的，如果出现逆序就不是 BST
func isValidBST1(root *TreeNode) bool {
arr := []int{}
inOrder(root, &arr)
for i := 1; i < len(arr); i++ {
if arr[i-1] >= arr[i] {
return false
}
}
return true
}

func inOrder(root *TreeNode, arr *[]int) {
if root == nil {
return
}
inOrder(root.Left, arr)
*arr = append(*arr, root.Val)
inOrder(root.Right, arr)
}

``````

Sep 6, 2020