classSolution(object):deffindMin(self, nums):""":type nums: List[int] :rtype: int"""foriinrange(len(nums) - 1):ifnums[i] > nums[i + 1]:returnnums[i + 1]returnnums[0] follow up: 数组中存在重复元素,处理方法与上一道题Search in Rotated Sorted Array一样,对边缘移动一步,直到边...
LeetCode 704. Binary Search (二分查找) 题目标签:Binary Search 很标准的一个二分查找,具体看code。 Java Solution: Runtime: 0 ms, faster than 100 % Memory Usage: 39 MB, less than 90 % 完成日期:07/31/2019 关键点:二分查找 classSolution {publicintsearch(int[] nums,inttarget) {intleft =...
welcome to my blog 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 to...
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...
输入:s = "010"输出:0解释:字符串已经是交替字符串了,不需要交换。 示例3: 输入:s = "1110"输出:-1 提示: 1 <= s.length <= 1000 s[i]的值为'0'或'1' 排序:最热 0 回复 © 2025 领扣网络(上海)有限公司 41 101 1 2 3 4 5 6 classSolution{ public: intminSwaps(strings) { } };...
102. 二叉树的层序遍历 - 给你二叉树的根节点 root ,返回其节点值的 层序遍历 。 (即逐层地,从左到右访问所有节点)。 示例 1: [https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg] 输入:root = [3,9,20,null,null,15,7] 输出:[[3],[9,20],[15,7]]
*/class Solution{public:TreeNode*bstToGst(TreeNode*root){intpre=0;order(root,pre);returnroot;}private:voidorder(TreeNode*root,int&pre){if(!root)return;stack<TreeNode*>s;while(root or s.size()){if(root){s.push(root);root=root->right;}else{root=s.top();s.pop();root->val+=pre...
Leetcode - Unique Binary Search Trees II My code: My test result: 这道题目也没做出来。但是看了解题思路后,觉得太强大了。这么短的十... Richardo92阅读 765评论 0赞 1 [Leetcode95]Unique Binary Search Trees II 和对递归怎... 题目描述 Given an integer n, generate all structurally unique......
A clever solution exists. Isn't this O(n2), though? It's also a leetcode problem.The in-order traversal gives you an ordering of the elements. You can reconstruct the original binary tree by adding elements to a binary search tree in the pre-order traversal order, with "<=>" ...
题目: Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? 题解: ...