/**
 * 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:

    int sum;
    vector<vector<int >> res;
    vector<int> path;
    vector<vector<int>> findPath(TreeNode* root, int sum) {
        //思路，深搜，如果搜索到叶子节点且sum == 0，当前路径合法，加入到res中，否则不合法，直接return
        //深搜中注意保护原来现场，所以有pop_back
        sum = sum;
        dfs(root,sum);
        return res;
    }

    void dfs(TreeNode* root,int sum){
        if(!root) return;
        path.push_back(root->val);
        sum -= root->val;
        if(!root->left && !root->right && !sum) res.push_back(path);
        dfs(root->left,sum);
        dfs(root->right,sum);
        path.pop_back();
    }
};