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->val
        backtrack(root->left,target-root->val);
        backtrack(root->right,target-root->val);
        path.pop_back();
    }
};