【LeetCode】113. 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.

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]
]

 

Solution 1

 1 class Solution {
 2 public:
 3     vector<vector<int>> pathSum(TreeNode* root, int sum) {
 4         vector<vector<int>> res;
 5         vector<int> cur;
 6         if (!root)
 7             return res;
 8         
 9         dfs(res, cur, root, sum);
10         return res;
11     }
12     
13     void dfs(vector<vector<int>>& res, vector<int>& cur, TreeNode* root, int sum) {
14         if (!root) {
15             return;
16         }
17         cur.push_back(root->val);
18         if (!root->left && !root->right && root->val == sum) {
19             
20             res.push_back(cur);
21             cur.pop_back();
22             return;
23         }
24         dfs(res, cur, root->left, sum - root->val);
25         dfs(res, cur, root->right, sum - root->val);
26         cur.pop_back();
27     }
28     
29 };

 

Solution 2

 1 class Solution {
 2 public:
 3     vector<vector<int>> pathSum(TreeNode* root, int sum) {
 4         vector<vector<int>> res;
 5         vector<TreeNode*> path;
 6         TreeNode* cur = root, *pre = nullptr;
 7         int val = 0;
 8         while (cur || !path.empty()) {
 9             while (cur) {
10                 path.push_back(cur);
11                 val += cur->val;
12                 cur = cur->left;
13             }
14             cur = path.back();
15             if (!cur->left && !cur->right && val == sum) {
16                 res.push_back({});
17                 for(auto it : path) {
18                     res.back().push_back(it->val);
19                 }
20             }
21             if (cur->right && cur->right != pre) 
22                 cur = cur->right;
23             else {
24                 pre = cur;
25                 val -= cur->val;
26                 path.pop_back();
27                 cur = nullptr;
28             }
29         }
30         return res;
31     }
32 };

 

posted @ 2018-04-14 10:32  Vincent丶丶  阅读(164)  评论(0)  编辑  收藏  举报