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 a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    void dfs_Path(TreeNode* node, int sum, int curSum, vector<int> path, vector<vector<int> > &pathSet){        if (node == NULL) return;        path.push_back(node->val);        curSum += node->val;        if(node->left == NULL && node->right == NULL){//到达叶节点            if(curSum == sum){//如果符合要求                pathSet.push_back(path);//保存            }            return;        }        dfs_Path(node->left, sum, curSum, path, pathSet);        dfs_Path(node->right, sum, curSum, path, pathSet);    }    vector<vector<int> > pathSum(TreeNode* root, int sum) {        vector<vector<int> > pathSet;        vector<int> path;        dfs_Path(root, sum, 0, path, pathSet);        return pathSet;    }};