/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ classSolution { public: boolisValidBST(TreeNode* root) { if(root==NULL)returntrue; returnisv(r...
classSolution {publicbooleanisValidBST(TreeNode root) {returnisValidBST(root, Integer.MIN_VALUE, Integer.MAX_VALUE); }publicbooleanisValidBST(TreeNode root,intminVal,intmaxVal) {if(root ==null) {returntrue; }if(root.val <= minVal || root.val >=maxVal) {returnfalse; }returnisValidBST(ro...
classSolution{publiclongval=Long.MIN_VALUE;publicbooleanisValidBST(TreeNode root){if(root ==null)returntrue;if(!isValidBST(root.left))returnfalse;if(val >= root.val)returnfalse; val = root.val;return isValidBST(root.right);}}/** * Definition for a binary tree node. * public class...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: bool isValidBST(TreeNode* root) { return judge(root,LONG_MIN,LONG_MAX); }...
https://leetcode.com/problems/validate-binary-search-tree/ 二叉树的问题,要想到递归。这里容易想到的就是如果左右子树都存在,只要 AI检测代码解析 if root.left.val < root.val < root.right.val: return self.isValidBST(root.left) and self.isValidBST(root.right) ...
id=98 lang=javascript * * [98] Validate Binary Search Tree *//** * Definition for a binary tree node. * function TreeNode(val) { * this.val = val; * this.left = this.right = null; * } *//** * @param {TreeNode} root * @return {boolean} */var isValidBST = functio...
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ func isValidBST(root *TreeNode) bool { return dfs(root, nil, nil) } func dfs(root *TreeNode, low *int, high *int) bool { // 空子树必定满足题意 if...
classSolution{publicbooleanisValidSerialization(String preorder){int n=preorder.length();int i=0;Deque<Integer>stack=newLinkedList<Integer>();stack.push(1);while(i<n){if(stack.isEmpty()){returnfalse;}if(preorder.charAt(i)==','){i++;}elseif(preorder.charAt(i)=='#'){int top=stack...
{ this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public boolean isValidBST(TreeNode root) { return isValidBST(root, null, null); } //定义:当前结点为根...
String[] args){ TreeNode root = new TreeNode(5); root.left = new TreeNode(1); root.right = new TreeNode(4); root.right.left = new TreeNode(3); root.right.right = new TreeNode(6); System.out.println(isValidBST(root)); }}【每日寄语】 生命不是用来寻找...