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...
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. Example 1: 2 / \ 1 3 1. 2. 3. Binary tree [2,1,3], return true. Example 2: 1 / \ ...
leetcode -- Validate Binary Search Tree -- 重点 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 if root.left.val < root.val < root.right.val: return self.isValidBST(root.left) and self.isValidBST(root.right)...
public boolean isValidBST(TreeNode root) { if (root == null) { return true; } boolean leftVailid = true; boolean rightVaild = true; if (root.left != null) { //大于左孩子并且左子树是合法二分查找树 leftVailid = root.val > root.left.val && isValidBST(root.left); } if (!left...
Leetcode 98. Validate Binary Search Tree Given a binary tree, determine if it is a valid binary se... ShutLove阅读 253评论 0赞 0 [leetcode] 98. Validate Binary Search Tree Given a binary tree, determine if it is a valid binary se... 叶孤陈阅读 399评论 0赞 0 LeetCode 98 Validate...
【Binary Tree】Validate Binary Search Tree 题目链接 https://leetcode.com/problems/validate-binary-search-tree/description/ 思路:左<根<右,中序遍历就行啦
Leetcode-Medium 98. Validate Binary Search Tree 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...
The above binary tree is serialized as{2,1,4,#,#,3,5}(in level order). Note 第一种,先跑左子树的DFS:如果不满足,返回false. 左子树分支判断完成后,pre = root.left(因为唯一给pre赋值的语句是pre = root,所以运行完dfs(root.left)之后,pre = root.left)。
虽然leetcode没有测试这个情况,不过健全的程序总是最好的。 class Solution { public: bool isValidBST(TreeNode *root) { return validBST(root); } //注意:别忘记了两边的boundary:leftMax和rightMax的设置 /* I dont think it's a good idea to use int to represent the up and low bound of a ...