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.

/**
 * 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 max(int a, int b)
    {
        return a > b? a : b;
    }

    int maxPathSum(TreeNode *root) {
        if (!root)
            return 0;
        int val,ll,rr;
        ll = rr = 0;
        val = -65535;
        if (root->left)
        {
            val = max(val,maxPathSum(root->left));
            ll = max(0,root->left->val);
        }
        if (root->right)
        {
            val = max(val,maxPathSum(root->right));
            rr = max(0,root->right->val);
        }
        val = max(val,root->val + ll + rr);
        root->val += max(ll,rr);
        return val;
    }
};

先计算左子节点到叶子节点最大值，在计算右子节点到叶子节点最大值，最后计算根节点最大值，val代表所求最大值

要更新当前节点到叶节点的最大路径值，考虑负数情况