LeetCode_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]
]
/**
* 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
- 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 ...
- 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 ...
- [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 ...
- 【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 ...
- 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 ...
- 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 ...
- 【leetcode】Combination Sum II
Combination Sum II Given a collection of candidate numbers (C) and a target number (T), find all uni ...
- [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 ...
- 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
- Hadoop:Task process exit with nonzero status of 1 异常
在运行hadoop程序时经常遇到异常 java.io.IOException: Task process exit with nonzero status of 1.网上很多博文都说是磁盘不够的问题. ...
- cf493D Vasya and Chess
D. Vasya and Chess time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- cf493A Vasya and Football
A. Vasya and Football time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
- html5 app图片预加载
function Laimgload(){} //图片预加载JS Laimgload.prototype.winHeight = function(){ //计算页面高度 var winHeight ...
- python3-day4(yield)
1.yield 迭代器是访问集合元素的一种方式.迭代器对象从集合的第一个元素开始访问,直到所有的元素被访问完结束.迭代器只能往前不会后退,不过这也没什么,因为人们很少在迭代途中往后退.另外,迭代器的一 ...
- Android ActionBar完全解析,使用官方推荐的最佳导航栏(下) .
转载请注明出处:http://blog.csdn.net/guolin_blog/article/details/25466665 本篇文章主要内容来自于Android Doc,我翻译之后又做了些加工 ...
- c++11 : range-based for loop
0. 形式 for ( declaration : expression ) statement 0.1 根据标准将会扩展成这样的形式: 1 { 2 auto&& __ra ...
- ASE中的主要数据库
Adaptive Server包括多种类型数据库: 必需数据库. “附加功能”数据库 .例子数据库 .应用数据库 1.必需数据库 master 数据库包含系统表,这些系统表中存储的数据被用来管理,有 ...
- IOS 原生解析JSON 问题
服务器----WebService 返回的是JSON数据 IOS解析报错: Error Domain=NSCocoaErrorDomain Code=3840 "Unable to conv ...
- C#基础:命令解析
1.普通格式命令的解析 例如: RENA<SP>E:\\A.txt<SP>C:\\B.txt<CRLF> (SP -> 空格,CRLF -> 回车加换行 ...