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[LeetCode]Binary Tree Maximum Path Sum

时间:2015-08-28 00:28:54      阅读:253      评论:0      收藏:0      [点我收藏+]

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

For example:
Given the below binary tree,

       1
      /      2   3

Return 6.

前序遍历,计算出最大值。

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     struct TreeNode *left;
 6  *     struct TreeNode *right;
 7  * };
 8  */
 9  
10 int preOrder(struct TreeNode* root, int *maxSum) {
11     int leftSum = 0, rightSum = 0, partSum = 0, sum = 0;
12     
13     if(!root) return 0;
14     
15     leftSum = preOrder(root->left, maxSum);
16     rightSum = preOrder(root->right, maxSum);
17     
18     partSum = root->val;
19     sum = rightSum + leftSum + root->val;
20     if(leftSum > rightSum && leftSum > 0) {
21         partSum += leftSum;
22     }
23     else if(rightSum > leftSum && rightSum > 0) {
24         partSum += rightSum;
25     }
26     if(*maxSum < partSum){
27         *maxSum = partSum;
28     }
29     if(*maxSum < sum){
30         *maxSum = sum;
31     }
32     
33     return partSum;
34 }
35 
36 int maxPathSum(struct TreeNode* root) {
37     if(!root) return 0;
38     
39     int maxSum = root->val;
40     preOrder(root, &maxSum);
41     
42     return maxSum;
43 }
92 / 92 test cases passed.
Status:

Accepted

Runtime: 20 ms

 

 

 

最后运行时间是20ms,但是看到统计图里有大神的代码只耗时16ms,看来是还有什么高级解法,希望看到本文的大神指点下啊!

[LeetCode]Binary Tree Maximum Path Sum

原文:http://www.cnblogs.com/zhangzhiqi/p/4765136.html

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