return dfs(root.Left, low, &root.Val) && dfs(root.Right, &root.Val, high) } 题目链接: Validate Binary Search Tree: leetcode.com/problems/v 带因子的二叉树: leetcode.cn/problems/va LeetCode 日更第 209 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满 ...
/* * @lc app=leetcode 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} *...
Both the left and right subtrees must also be binary search trees. 一开始是这样做的: 1classSolution {2public:3boolisValidBST(TreeNode*root) {4returncheckValid(root, INT_MIN, INT_MAX);5}67boolcheckValid(TreeNode * root,intleft,intright)8{9if(root ==NULL)10returntrue;11return(root->...
[LeetCode] 98. Validate Binary Search Tree 验证二叉搜索树 验证二叉搜索树CategoryDifficultyLikesDislikes algorithms Medium (31.28%) 620 - TagsCompanies给定一个二叉树,判断其是否是一个有效的二叉搜索树。假设一个二叉搜索树具有如下特征: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...
LeetCode-98. Validate Binary Search Tree 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 keysless thanthe node's key....
Both the left and right subtrees must also be binary search trees. confused what"{1,#,2,3}"means?> read more on how binary tree is serialized on OJ. 这道验证二叉搜索树有很多种解法,可以利用它本身的性质来做,即左<根<右,也可以通过利用中序遍历结果为有序数列来做,下面我们先来看最简单的...
Leetcode-Medium 98. Validate Binary Search Tree 题目描述 判定一棵树是否满足二叉搜索树的性质。二叉查找树(Binary Search Tree),(又:二叉搜索树,二叉排序树)它或者是一棵空树,或者是具有下列性质的二叉树: 若它的左子树不空,则左子树上所有结点的值均小于它的根结点的值; 若它的右子树不空,则右子树上所有...
* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */classSolution{publicbooleanisValidBST(TreeNoderoot){Integermin=null;Integermax=null;returnisValid(root,min,max);}publicbooleanisValid...
Leetcode 98. Validate Binary Search Tree验证二叉搜索树 MaRin 菜鸡一只二叉搜索树的性质是,左子树上所有的元素都要比根节点小,右子树上所有的元素都要比根节点大,那么按照这个性质,第一种思路就是从根节点开始中序遍历,得到一个升序的序列即为二叉搜索树 思路1:判断中序遍历是否升序,这里不需要把所有元素都遍历...
# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None # 方法一 class Solution(object): def isValidBST(self, root): """ :type root: TreeNode :rtype: bool """ res = [] self.in...