×【leetcode】Binary Tree Maximum Path Sum

链接：https://oj.leetcode.com/problems/binary-tree-maximum-path-sum/

描述：

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.

解法：

递归求解，

每一个递归返回两个值，一个值是以该节点为根的树中的最大和，另一个值是一该节点为某一起始点的最大路径。

此题思路需要十分清晰，容易出错。另注意树中含有负数的情况。

代码如下：

    int maxPathSum(TreeNode *root) {    if(root == NULL) return 0;    if(root->left == NULL && root->right == NULL) return root->val;    int result = 0;    maxSum(root, result);    return result;    }        int maxSum(TreeNode* root, int &maxsum)    {    if( root == NULL) {    maxsum = INT_MIN;    return 0;    }        int leftsum = INT_MIN, rightsum = INT_MIN;    int left = maxSum(root->left, leftsum);    int right = maxSum(root->right, rightsum);    int sum = root->val;    if( left > 0)    sum += left;    if(right > 0)    sum += right;    maxsum = max(leftsum, rightsum);    maxsum = max(maxsum, sum);    return max(left, right) > 0 ? max(left, right) + root->val : root->val;;    }