「分治法」Binary Tree Maximum Path Sum

/*************************************************************************
	> File Name: BinaryTreeMaximumPathSum.cpp
	> Author: Shaojie Kang
	> Mail: [email protected] 
	> Created Time: 2015年09月16日 星期三 16時06分51秒 
    > Problem:
        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. 
    > Solution:
        divide and conquer
 ************************************************************************/

#include<iostream>
#include<limits.h>
using namespace std;

class TreeNode 
{
public:
    int val;
    TreeNode *left, *right;
    TreeNode(int val)
    {
        this->val = val;
        this->left = this->right = NULL;
    }
};

class Solution 
{
public:
    int maxPathSum(TreeNode *root)
    {
        int maxSum = INT_MIN;
        maxPathSumCore(root, maxSum);
        return maxSum;
    }

    int maxPathSumCore(TreeNode *root, int &maxSum)
    {
        if(root == NULL) return 0;
        int sum = root->val;
        int leftSum = maxPathSumCore(root->left, maxSum);
        int rightSum = maxPathSumCore(root->right, maxSum);
        if(leftSum > 0) sum += leftSum;
        if(rightSum > 0) sum += rightSum;
        if(sum > maxSum) maxSum = sum;
        return max(leftSum, rightSum) > 0 ? root->val + max(leftSum, rightSum) : root->val;
    }
};