#94 Binary Tree Maximum Path Sum

题目描述：

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

Have you met this question in a real interview? 

Yes

Example

Given the below binary tree:

  1 / \2   3

return 6.

题目思路：

这题因为path可以起始/终止于任何一个node，似乎不太好下手。但是可以发现，无论怎样，path都会经过某些node。那如果对于某一node，它的左子树（包含左节点）出一条path（这条path必须不能同时包含左子树的左右子树），右子树（包含右节点）出一条path，那么max可能是左path，或者右path，或者只是这个node本身，或者是左path+node+右path。

Mycode（AC = 51ms）：

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */class Solution {public:    /**     * @param root: The root of binary tree.     * @return: An integer     */    int maxPathSum(TreeNode *root) {        // write your code here        if (root == NULL) {            return 0;        }                int max_sum = INT_MIN;        maxPathSum(root, max_sum);        return max_sum;    }        int maxPathSum(TreeNode *root, int& max_sum) {        if (!root->left && !root->right) {            max_sum = max(max_sum, root->val);            return root->val;        }                int left = INT_MIN, right = INT_MIN, result = INT_MIN;                // get the max path from left subtree (include left node)        if (root->left) {            left = maxPathSum(root->left, max_sum);            result = max(root->val, root->val + left);        }                // get the max path from right subtree (include right node)        if (root->right) {            right = maxPathSum(root->right, max_sum);            result = max(result, max(root->val, root->val + right));        }                if (left != INT_MIN && right != INT_MIN) {            max_sum = max(left + right + root->val, max(max_sum, result));        }        else {            max_sum = max(max_sum, result);        }                return result;    }};