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:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int> > res;
        vector<int> solution;
        dfs(root, sum, solution, res);
        return res;
    }
private:
    void dfs(TreeNode *r,int gap, vector<int> &path, vector<vector<int> > &ret){
        if(r == NULL){
            //不在这里push_back的原因知道吗，是因为此时
            //如果叶子节点满足条件了，会继续遍历叶子节点的
            //两个孩子节点，都是NULL,这样就会重复
            //if(gap == 0)    ret.push_back(path);
            return;
        }

        path.push_back(r->val);
        if(r->left == NULL && r->right == NULL){
            if(gap == r->val)
                ret.push_back(path);
        }


        //只有到达NULL节点才会返回
        dfs(r->left, gap-r->val, path, ret);

        dfs(r->right, gap-r->val, path, ret);

        path.pop_back();
    }
};