题目: Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1 return…

[LeetCode]113. Path Sum II 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/path-sum-ii/description/ 题目描述: Given a binary tree and a sum, find all root-to-leaf paths where each…

Minimum Path Sum Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time.   动态规划即可,与Unique…

problem 437. Path Sum III 参考 1. Leetcode_437. Path Sum III; 完…

Path Sum II Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return [ [5,4,11,2], [5,8,4,5] ] 这题是在Path…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 DFS 日期 题目地址:https://leetcode-cn.com/problems/path-sum-iv/ 题目描述 If the depth of a tree is smaller than 5, then this tree can be represented by a list of three-digits integers. Fo…

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作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 DFS + DFS BFS + DFS 日期 题目地址:https://leetcode.com/problems/path-sum-iii/#/description 题目描述 You are given a binary tree in which each node contains an integer value. Find the num…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 文章目录 题目描述 题目大意 解题方法 BFS DFS 日期 题目地址:https://leetcode.com/problems/path-sum-ii/description/ 题目描述 Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given…

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time. 思路:由于只能向两个方向走,瞬间就没有了路线迂回的烦恼,题目的难度…