Leetcode -- Binary Tree Maximum Path Sum

Given a binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path does not need to go through the root.

For example:

Given the below binary tree,

1
/ \
2   3

Return

6

.

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int maxsum=INT_MIN;
int dfs(TreeNode* root)
{
if(root==NULL) return 0;
int left = dfs(root->left);
int right = dfs(root->right);
int sum = root->val;
if(left>0) sum+=left;
if(right>0) sum+= right;
maxsum = max(maxsum,sum);
return max(left,right)>0?max(left,right)+root->val:root->val;
}
int maxPathSum(TreeNode* root) {
dfs(root);
return maxsum;
}
};

分析：

此题中dfs返回的值并不是最终结果，只是单方向路径，而最终的最大和在遍历过程中被计算。

在dfs的过程中，以每个节点为中心计算一次sum，保存其最大值，计算sum的过程中需要左右路径的单方向求和，所以返回此单方向路径和。