LeetCode113 Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.（Medium）

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

分析：

回溯法处理即可，利用helper函数遍历所有路径，达到sum时添加到result里。

代码：

 /**
  * 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 {
     vector<vector<int>> result;
     void helper(TreeNode* root, int sum, vector<int>& vec) {
         if (root == nullptr) {
             return;
         }
         vec.push_back(root -> val);
         sum -= root -> val;
         if (root -> left == nullptr && root -> right == nullptr) {
             if (sum == ) {
                 result.push_back(vec);
             }
             else {
                 vec.pop_back();
                 sum += root -> val;
                 return;
             }
         }
         if (root -> left != nullptr) {
             helper(root -> left, sum, vec);
         }
         if (root -> right != nullptr) {
             helper(root -> right, sum, vec);
         }
         vec.pop_back();
         sum -= root -> val;
     }
 public:
     vector<vector<int>> pathSum(TreeNode* root, int sum) {
         vector<int> vec;
         helper(root, sum, vec);
         return result;
     }
 };