root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8 返回3。和等于 8 的路径有: 1. 5 -> 3 2. 5 -> 2 -> 1 3. -3 -> 11 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/path-sum-iii 常规思路:双递归,先序遍历的同时计算每个节点的路径和数量。 1 2 3 4 5 ...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ classSolution{ public: vector<int>distanceK(TreeNode*root,TreeNode*target,intk) { ...
https://leetcode.com/problems/all-nodes-distance-k-in-binary-tree/ https://leetcode.com/problems/all-nodes-distance-k-in-binary-tree/discuss/143752/JAVA-Graph-%2B-BFS https://leetcode.com/problems/all-nodes-distance-k-in-binary-tree/discuss/143775/very-easy-to-understand-c%2B%2B-solution...
Can you solve this real interview question? All Nodes Distance K in Binary Tree - Given the root of a binary tree, the value of a target node target, and an integer k, return an array of the values of all nodes that have a distance k from the target node
560. 和为 K 的子数组 - 给你一个整数数组 nums 和一个整数 k ,请你统计并返回 该数组中和为 k 的子数组的个数 。 子数组是数组中元素的连续非空序列。 示例 1: 输入:nums = [1,1,1], k = 2 输出:2 示例 2: 输入:nums = [1,2,3], k = 3 输出:2 提示
1679. Max Number of K-Sum Pairs # 题目 # You are given an integer array nums and an integer k. In one operation, you can pick two numbers from the array whose sum equals k and remove them from the array. Return the maximum number of operations you can
leetcode-863-All Nodes Distance K in Binary Tree,Converttreetoindirectgraph,thenuseBFStogetKdistancefromtarget.Error:DonotfullyunderstandtheBFS,whenwantstogetsmallestdistance
3.1 Segment Tree 3.2 UnionFind 3.3 LRUCache 3.4 LFUCache 3.5 Binary Indexed Tree 第四章 LeetCode 题解 0001~0099 0001. Two Sum 0002. Add Two Numbers 0003. Longest Substring Without Repeating Characters 0004. Median of Two Sorted Arrays 0005. Longest Palindromic Substring 0006. Zig Zag Conversio...
My Solutions to Leetcode problems. All solutions support C++ language, some support Java and Python. Multiple solutions will be given by most problems. Enjoy:) 我的Leetcode解答。所有的问题都支持C++语言,一部分问题支持Java语言。近乎所有问题都会提供多个算
n self.tree[idx] = value while idx > 1: idx //= 2 self.tree[idx] = self.tree[idx * 2] + self.tree[idx * 2 + 1] def query(self, left, right): # Query the sum in [left, right) result = 0 left += self.n right += self.n while left < right: if left & 1: ...