维护sum变量，当达到叶子节点时且sum变为0的时候，找到合适的路径
遍历左右子树，遍历完毕的时候path需要pop恢复path
/**
 * 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>> ans;
    vector<int> path;

    vector<vector<int>> findPath(TreeNode* root, int sum) {
        dfs(root, sum);
        return ans;
    }

    void dfs(TreeNode* root, int sum){
        if(!root) return ;         //空节点肯定不是叶子节点
        sum -= root -> val;
        path.push_back(root -> val);
        if(!root -> left && !root -> right && sum == 0)  ans.push_back(path);
        dfs(root -> left, sum);
        dfs(root -> right, sum);
        path.pop_back();
    }
};