1.题目描述 2.问题分析 对于每个节点,如果其左子节点是叶子,则加上它的值,如果不是,递归,再对右子节点递归即可. 3.代码 int sumOfLeftLeaves(TreeNode* root) { if (root == NULL) ; ; if (root->left != NULL) { if (root->left->left == NULL && root->left->right == NULL) ans += root->left-&g…

404. Sum of Left Leaves [题目]中文版  英文版 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int sumOfLeftLea…

前言   [LeetCode 题解]系列传送门:  http://www.cnblogs.com/double-win/category/573499.html   1.题目描述 Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The same repeated number…

前言   [LeetCode 题解]系列传送门:  http://www.cnblogs.com/double-win/category/573499.html   1.题目描述 Given an array of integers, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers suc…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目大意 题目大意 解题方法 递归 迭代 日期 [LeetCode] 题目地址:https://leetcode.com/problems/sum-of-left-leaves/ Difficulty: Easy 题目大意 Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \…

题目地址:https://oj.leetcode.com/problems/two-sum/ Two Sum Given an array of integers, find two numbers such that they add up to a specific target number. The function twoSum should return indices of the two numbers such that they add up to the target, w…

这是悦乐书的第217次更新,第230篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第85题(顺位题号是404).找到给定二叉树中所有左叶的总和.例如: 二叉树中有两个左叶,分别为9和15. 返回24. 3 / \ 9 20 / \ 15 7 本次解题使用的开发工具是eclipse,jdk使用的版本是1.8,环境是win7 64位系统,使用Java语言编写和测试. 02 第一种解法 使用递归. 特殊情况:当root为null时,直接返回0. 正常情况:定义一个变量su…

------------------------------------------------------------------- 分两种情况: 1.当前节点拥有左孩子并且左孩子是叶子节点:左孩子值+右孩子子树遍历统计2.不符合上面那种情况的从当前节点劈开为两颗子树分别统计相加即可 AC代码: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * Tre…

1.题目描述 2.问题分析 记录所有路径上的值,然后转换为int求和. 3.代码 vector<string> s; int sumNumbers(TreeNode* root) { traverse(root, ""); ; for(auto it = s.begin(); it != s.end(); it++) { sum += stoi(*it); } return sum; } void traverse(TreeNode *t, string str) { if…

#-*- coding: UTF-8 -*- # Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):        def dfs(self,root,isLeft):      …