/**
 * 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>> res;                        //储存最终结果
    vector<int> path;                               //储存每条路径的结果

    void dfs(TreeNode* root, int sum) {
        if (!root) return;

        sum -= root->val;
        path.push_back(root->val);

        if (!root->left && !root-> right && !sum)   //遍历到叶子节点且和相等时
            res.push_back(path);

        dfs(root->left, sum);       //这里不用判空，因为函数第一行已经判空了
        dfs(root->right, sum);
        path.pop_back();            //恢复现场，因为sum是局部变量，故无须恢复
    }

    vector<vector<int>> findPath(TreeNode* root, int sum) {
        if (!root) return res;
        dfs(root, sum);

        return res;
    }
};