LeetCode 129. Sum Root to Leaf Numbers

原题链接在这里：https://leetcode.com/problems/sum-root-to-leaf-numbers/

题目：

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   / \
  2   3

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

题解：

Top-down dfs到了叶子节点加值.

Time Complexity: O(n).

Space: O(h), h是树的高度.

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     public int sumNumbers(TreeNode root) {
12         if(root == null){
13             return 0;
14         }
15         
16         int [] res = {0};
17         dfs(root, 0, res);
18         return res[0];
19     }
20     
21     private void dfs(TreeNode root, int cur, int [] res){
22         if(root == null){
23             return;
24         }
25         
26         cur = cur*10+root.val;
27         
28         if(root.left == null && root.right == null){
29             res[0] += cur;
30             return;
31         }
32 
33         dfs(root.left, cur, res);
34         dfs(root.right, cur, res);
35 
36     }
37 }

AC C++:

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 8  *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 9  *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10  * };
11  */
12 class Solution {
13 public:
14     int sumNumbers(TreeNode* root) {
15         int res = 0;
16         dfs(root, 0, res);
17         return res;
18     }
19 
20 private:
21     void dfs(TreeNode* root, int cur, int& res){
22         if(!root){
23             return;
24         }
25 
26         cur = cur * 10 + root->val;
27         if(!root->left && !root->right){
28             res += cur;
29         }
30 
31         dfs(root->left, cur, res);
32         dfs(root->right, cur, res);
33     }
34 };

类似Path Sum II.

posted @ 2015-09-07 22:29  Dylan_Java_NYC  阅读(256)  评论(0)  编辑  收藏  举报