leetcode: Binary Tree Maximum Path Sum

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

.

考虑经过根节点和不经过根节点两种情况

/**
* 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 maxi = INT_MIN;
core( root, maxi);
return maxi;
}
int core( TreeNode *root, int &maxi){
if( root == NULL)
return 0;
int leftsum = core( root->left, maxi);
int rightsum = core( root->right, maxi);
//经过自己，连通左右子树
int rootsum = root->val;
rootsum = leftsum > 0 ? leftsum + rootsum : rootsum;//左子树大于0就加上
rootsum = rightsum > 0 ? rightsum + rootsum : rootsum;//右子树大于0就加上
maxi = rootsum > maxi ? rootsum : maxi;
//不连通左右子树
int pathsum = root->val;
int subsum = max( leftsum, rightsum);
pathsum = subsum > 0 ? pathsum + subsum : pathsum;
maxi = pathsum > maxi ? pathsum : maxi;

return pathsum;

}
};