思路：利用回溯的思想，回溯的退出条件为当前节点为空，是符合路径的判断条件为路径和为目标值且叶子节点包含了，代码如下：

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:vector<vector<int>> res;vector<int> path;vector<vector<int>> pathTarget(TreeNode* root, int target) {backtrack(root,target);return res;}void backtrack(TreeNode* root, int target){if(root == nullptr) return;//路径写入当前节点的值path.push_back(root->val);//如果当前节点加上之后等于target且当前节点为叶子节点则将当前路径写入结果中if(target - root->val ==0 &&root->left == nullptr &&root->right == nullptr){res.push_back(path);}//向左右节点回溯//如果root为叶节点则不回溯了，并且路径中弹出root->valbacktrack(root->left,target-root->val);backtrack(root->right,target-root->val);path.pop_back();}
};