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LeetCode OJ:Sum Root to Leaf Numbers(根到叶节点数字之和)

时间:2015-10-26 12:08:17      阅读:217      评论:0      收藏:0      [点我收藏+]

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.

简单的用递归就可以实现,代码如下:

 1 class Solution {
 2 public:
 3     int sumNumbers(TreeNode* root) {
 4         if(!root) return 0;
 5         dfs(root, 0);
 6     }
 7 
 8     int dfs(TreeNode * root, int curr){
 9         if(!root->left && !root->right) return curr;
10         if(!root->left && root->right) return dfs(root->right, curr * 10);
11         if(root->left && !root->right) return dfs(root->left, curr * 10);
12         else return dfs(root->left, curr * 10) + dfs(root->right, curr * 10);
13     }
14 private:
15     int sum;
16 };

 

LeetCode OJ:Sum Root to Leaf Numbers(根到叶节点数字之和)

原文:http://www.cnblogs.com/-wang-cheng/p/4910721.html

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