Binary Tree Maximum Path Sum

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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.

 

For each node like following, there should be four ways existing for max path:

1. Node only
2. L-sub + Node
3. R-sub + Node
4. L-sub + Node + R-sub

 

注意：函數返回時返回的是left+node, node, right+node中較大的值。而計算全域最大是比較以當前node為根的樹和（left+right+node,node, left+node,right+node）之前記錄的最大值。

 

 1 /** 2  * Definition for binary tree 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     int maxPathSum(TreeNode *root) {13         if (root==NULL)return 0;14         int max_sum=INT_MIN;15         pathSum(root,max_sum);16         return max_sum;17         18     }19     int pathSum(TreeNode* root, int& max_sum)20     {21         if(root==NULL)return 0;22         int sum=root->val;23         int left=pathSum(root->left,max_sum);24         sum=max(sum,sum+left);25         int right=pathSum(root->right,max_sum);26         sum=max(sum,sum+right);27         max_sum=max(sum,max_sum);28         return max(root->val,max(left+root->val,right+root->val));//important29     }30 };