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 Binary tree[2,1,3], return true. Example 2: 1 /\2 3 Binary tree[1,2,3], return false. 2、边界条件...
题目链接: Binary Search Tree Iterator : leetcode.com/problems/b 二叉搜索树迭代器: leetcode-cn.com/problem LeetCode 日更第 95 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满 发布于 2022-04-25 09:24 力扣(LeetCode) Python 算法 赞同添加评论 分享喜欢收藏申请转载 ...
e. 使用DFS+stack实现:(参考:https://leetcode.wang/leetCode-98-Validate-Binary-Search-Tree.html) 1publicbooleanisValidBST(TreeNode root) {2if(root ==null|| root.left ==null&& root.right ==null) {3returntrue;4}5//利用三个栈来保存对应的节点和区间6LinkedList<TreeNode> stack =newLinkedLis...
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 ...
125 -- 1:40 App LeetCode 每日一题 Daily Challenge 704 Binary Search 372 -- 3:35 App LeetCode 每日一题 Daily Challenge 1305 All Elements in Two Binary Search Trees 188 1 1:34 App LeetCode 每日一题 Daily Challenge 669 Trim a Binary Search Tree 189 -- 4:44 App LeetCode 每日一题 ...
leetcode 669. Trim a Binary Search Tree 修建二叉搜索树BST + 深度优先遍历DFS,Givenabinarysearchtreeandthelowestandhighestboundariesas
You are given therootof a binary search tree (BST) and an integerval. Find the node in the BST that the node's value equalsvaland return the subtree rooted with that node. If such a node does not exist, returnnull. Example 1: ...
[LeetCode] 96. Unique Binary Search Trees n, how many structurally unique BST's (binary search trees) that store values 1 ...n? Example: Input: 3 Output: 5 Explanation: Givenn= 3, there are a total of 5 unique BST's: 1 3 3 2 1...
今天介绍的是LeetCode算法题中Easy级别的第163题(顺位题号是700)。给定一个二叉搜索树(BST)的和正整数val。 你需要在BST中找到节点的值等于给定val的节点。返回以该节点为根的子树。如果此节点不存在,则应返回null。例如: 鉴于树: 4 / \ 2 7 / \ 1 3 ...
Insert into a Binary Search Tree 2. Solution Iterative /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} ...