二叉树中和为某一值的路径
    题目链接：https://leetcode-cn.com/problems/er-cha-shu-zhong-he-wei-mou-yi-zhi-de-lu-jing-lcof/
    图片.png

    思路：递归，每次从target中减去当前节点的值，终止条件为判断到达叶子节点时target是否为0. 返回上层时把当前节点的值要pop出去。

    参考代码：

    1. class Solution {
    2. public:
    3.     vector<vector<int>> pathSum(TreeNode* root, int target) {
    4.         vector<vector<int>> res;
    5.         if (!root) {
    6.             return res;
    7.         }
    8.         vector<int> path;
    9.         recursionSum(root, target, res, path);
    10.         return res;
    11.     }
    13.     void recursionSum(TreeNode* root, int target, vector<vector<int>>& res, vector<int>& path) {
    14.         if (!root) {
    15.             return;
    16.         }
    17.         path.push_back(root->val);
    18.         target -= root->val;
    19.         if (!root->left && !root->right) {
    20.             if (target == 0) {
    21.                 res.push_back(path);
    22.             }
    23.         }
    24.         recursionSum(root->left, target, res, path);
    25.         recursionSum(root->right, target, res, path);
    26.         path.pop_back();
    27.         return;
    28.     }
    29. };