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.
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int maxPathSum(TreeNode *root) {
if (root == NULL)
return 0;
int result = INT_MIN;
int maxRoot = maxPathSum(root, result);
return result;
}
int maxPathSum(TreeNode *root,int& result){
if (root == NULL)
return 0;
int left = maxPathSum(root->left, result);
int right = maxPathSum(root->right, result);
int temp;
int maxVal = max(left,right);
temp = max(maxVal+root->val,root->val);
temp = max(temp, left+right+root->val);
if (temp>result)
result = temp;
if (maxVal > 0)
return maxVal+root->val;
else
return root->val;
}
};
1. 注意比較最大值的時候,四種情況都要考慮到:
只有一個點,左子樹+點,右子樹+點,左子樹+右子樹+點
2, result爲包括某一點在內的,左右路徑的最大值