二叉树中和为某一值的路径

/**
 * 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:
    void pathSum(TreeNode* root, int target, std::vector<int>& tmp) {
      if (root == NULL) {
          return;
      }
      int val = root->val;
      int t1 = target - val;
      tmp.push_back(val);
      if (t1 == 0) {
          if (!root->left && !root->right) {
              res.push_back(tmp);
          }
      }
      pathSum(root->left, t1, tmp);
      pathSum(root->right, t1, tmp);
      tmp.erase(tmp.end() - 1);  // 删除最末尾元素，这样形参可以传递引用，减少内存开销
    }
    vector<vector<int>> pathSum(TreeNode* root, int target) {
        std::vector<int> tmp;
        pathSum(root, target, tmp);
        return res;
    }
    std::vector<std::vector<int>> res;
};