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 ...
返回一棵 BST的两个结点的最小绝对值的距离之差。 Given a binary search tree with non-negative values, find the minimumabsolute differencebetween values of any two nodes. 题解:根据 BST 的性质,我们只需要用一个变量记录中序遍历的前一个结点prev即可。 View Code 【683】K Empty Slots 【699】Falling...
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...
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 ...
https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/ 中文版描述 给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。百度百科中最近公共祖先的定义为:“对于有根树 T 的两个结点 p、q,最近公共...
The number of nodes in the tree is in the range[1, 5000]. 1 <= Node.val <= 107 rootis a binary search tree. 1 <= val <= 107 Copyright ©️ 2025 LeetCode All rights reserved 6.2K 64 0 Online Case 1 Case 2 [4,2,7,1,3] ...
LeetCode Top 100 Liked Questions 96. Unique Binary Search Trees (Java版; Medium) 题目描述 Given n, how many structurally unique BST's (binary search trees) that store values 1 ...n? Example: Input: 3 Output: 5 Explanation: Given n = 3, there are a total of 5 unique BST's: ...
leetcode 669. Trim a Binary Search Tree 修建二叉搜索树BST + 深度优先遍历DFS,Givenabinarysearchtreeandthelowestandhighestboundariesas
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) {} ...