LintCode 376: Binary Tree Path Sum

Description:
Given a binary tree, find all paths that sum of the nodes in the path equals to a given number target.


Note:
1.a valid path的定义是 from root node to any of the leaf nodes.所以当满足target条件时还要判断最后一个结点的左右子树是否为空（啊每次都不认真审题要注意啊）
2.注意函数getSum中的sum和path是不加引用，而result是加引用的，因为sum和path在递归中是不共享的，每次更新只作为参数传递下去，而result需要在递归中共享结果。

Code:

class Solution {void getSum(TreeNode* root, int target, int sum, vector<int> path, vector<vector<int>>& result) {if (!root)return;if (root->val + sum == target && !root->left && !root->right) {path.push_back(root->val);result.push_back(path);return;}sum += root->val;path.push_back(root->val);getSum(root->left, target, sum, path, result);getSum(root->right, target, sum, path, result);return;}public:/*** @param root the root of binary tree* @param target an integer* @return all valid paths*/vector<vector<int>> binaryTreePathSum(TreeNode *root, int target) {// Write your code herevector<vector<int>> result;vector<int> path;int sum = 0;getSum(root, target, sum, path,result);return result;}};