leecode 解题总结：113. Path Sum II

#include <iostream>#include <stdio.h>#include <vector>using namespace std;/*问题: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   1return[   [5,4,11,2],   [5,8,4,5]]分析:这和之前的要求不一样，需要输出从根节点到叶节点的值，我们可以用一个数组存储结点的值。具体做法如下:存储当前结点的值到数组中，如果当前结点是叶子结点，计算数组中所有元素的和是否等于给定值，如果等于，就将数组存储到结果集中；否则，表明当前叶子结点所有值不符合，从数组中弹出当前叶子结点的值，关键:1 //注意result不能用引用，会改变另一条路径的值void dfs(TreeNode* root , int sum , vector<int> result , vector< vector<int> >& results)*/struct TreeNode {    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {public://注意result不能用引用，会改变另一条路径的值void dfs(TreeNode* root , int sum , vector<int> result , vector< vector<int> >& results){//当前结点为空，肯定不可能，直接返回if(!root){return ;}result.push_back(root->val);sum -= root->val;//如果当前结点为叶子结点，计算结果if(NULL == root->left && NULL == root->right){//如果符合要求if(0 == sum){results.push_back(result);}//不符合要求，需要弹出当前结点对应的值else{result.pop_back();}}//如果当前结点的左孩子不空，递归处理if(root->left){dfs(root->left , sum , result , results);}if(root->right){dfs(root->right , sum , result , results);}}    vector<vector<int>> pathSum(TreeNode* root, int sum) {        vector<vector<int>> results;vector<int> result;dfs(root , sum , result , results);return results;    }};void print(vector<int>& result){if(result.empty()){cout << "no result" << endl;return;}int size = result.size();for(int i = 0 ; i < size ; i++){cout << result.at(i) << " " ;}cout << endl;}void process(){ vector<int> nums; int value; int num; Solution solution; vector<int> result; while(cin >> num ) { nums.clear(); for(int i = 0 ; i < num ; i++) { cin >> value; nums.push_back(value); } }}int main(int argc , char* argv[]){process();getchar();return 0;}