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]
]

  

/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void DFS(TreeNode *root, vector<int> &temp, int tempSum)
{
if(root->left == NULL && root -> right== NULL && tempSum == sum) {
result.push_back(temp);
return;
} if(root->left!= NULL ){
temp.push_back(root->left->val);
DFS(root->left, temp,root->left->val + tempSum );
temp.pop_back();
} if(root->right != NULL ){
temp.push_back(root->right->val);
DFS(root->right, temp, root->right->val + tempSum);
temp.pop_back();
}
}
vector<vector<int> > pathSum(TreeNode *root, int sum) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
result.clear();
if(!root) return result;
this->sum = sum;
vector<int> temp;
temp.push_back(root->val);
DFS(root, temp, root->val);
return result;
}
private:
int sum;
vector<vector<int>> result;
};

LeetCode_Path Sum II的更多相关文章

  1. Leetcode 笔记 113 - Path Sum II

    题目链接:Path Sum II | LeetCode OJ Given a binary tree and a sum, find all root-to-leaf paths where each ...

  2. Path Sum II

    Path Sum II Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals ...

  3. [leetcode]Path Sum II

    Path Sum II Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals ...

  4. 【leetcode】Path Sum II

    Path Sum II Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals ...

  5. 32. Path Sum && Path Sum II

    Path Sum OJ: https://oj.leetcode.com/problems/path-sum/ Given a binary tree and a sum, determine if ...

  6. LeetCode: Path Sum II 解题报告

    Path Sum II Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals ...

  7. 【leetcode】Combination Sum II

    Combination Sum II Given a collection of candidate numbers (C) and a target number (T), find all uni ...

  8. [LeetCode] #167# Two Sum II : 数组/二分查找/双指针

    一. 题目 1. Two Sum II Given an array of integers that is already sorted in ascending order, find two n ...

  9. leetcode2 Two Sum II – Input array is sorted

    Two Sum II – Input array is sorted whowhoha@outlook.com Question: Similar to Question [1. Two Sum], ...

随机推荐

  1. Linux企业级项目实践之网络爬虫(1)——项目概述及准备工作

    我们在学习了Linux系统编程之后,需要一些实战项目来提高自己的水平,本系列我们通过编写一个爬虫程序,将我们学习的知识进行综合应用,同时在实现项目的过程中逐渐养成一些有用的思维方式,并具有初步的软件开 ...

  2. 一位搬家师傅的O2O之旅

    一位搬家师傅的O2O之旅 By 诸神之黄昏 | 2014/08/14 [核心提示] 一位普通的搬家师傅,无意中被卷入如火如荼的 O2O 浪潮,起初,互联网让他尝到了甜头,后来则是更多的困惑和不解. 再 ...

  3. Maven--生命周期和插件(四)

    <Maven--搭建开发环境(一)> <Maven--构建企业级仓库(二)> <Maven—几个需要补充的问题(三)> <Maven—生命周期和插件(四)&g ...

  4. Entify Framewrok - Join的使用方法

    问题:有2个表,使用id相连,如何用Join语法将其连接起来? 如下代码 List<tblAssociation> assoList = dataContext.tblAssociatio ...

  5. Hybrid App开发模式中, IOS/Android 和 JavaScript相互调用方式

    IOS:Objective-C 和 JavaScript 的相互调用 iOS7以前,iOS SDK 并没有原生提供 js 调用 native 代码的 API.但是 UIWebView 的一个 dele ...

  6. iOS开发:深入理解GCD 第一篇

    最近把其他书籍都放下了,主要是在研究GCD.如果是为了工作,以我以前所学的GCD.NSOperation等知识已经足够用了,但学习并不仅仅知识满足于用它,要知其然.并且知其所以然,这样才可以不断的提高 ...

  7. HDU 3697 Selecting courses(贪心)

    题目链接:pid=3697" target="_blank">http://acm.hdu.edu.cn/showproblem.php?pid=3697 Prob ...

  8. C++——try、throw、catch实例学习程序

    #include<iostream> #include<stdexcept> //exception/stdexcept/new/type_info头文件里都有定义的标准异常类 ...

  9. IT English Collection(9) of Objective-C

    1 前言 今天我们来解除一篇有关Objective-C的介绍文章,详情如下. 2 详述 2.1 原文 Objective-C defines a small but powerful set of e ...

  10. C++11 静态断言(static_assert)

      简介 C++0x中引入了static_assert这个关键字,用来做编译期间的断言,因此叫做静态断言. 其语法很简单:static_assert(常量表达式,提示字符串). 如果第一个参数常量表达 ...