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.

Ref:http://fisherlei.blogspot.com/2013/01/leetcode-binary-tree-maximum-path-sum.html

[Thoughts]
For each node like following, there should be four ways existing for max path:

1. Node only
2. L-sub + Node
3. R-sub + Node
4. L-sub + Node + R-sub

Keep trace the four path and pick up the max one in the end.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int maxPathSum(TreeNode root) {
        int max[] = new int[1]; 
        max[0] = Integer.MIN_VALUE;
        calculateSum(root, max);
        return max[0];
    }
 
    public int calculateSum(TreeNode root, int[] max) {
        if (root == null)
            return 0;
 
        int left = calculateSum(root.left, max);
        int right = calculateSum(root.right, max);
 
        int current = Math.max(root.val, Math.max(root.val + left, root.val + right));
 
        max[0] = Math.max(max[0], Math.max(current, left + root.val + right));
 
        return current;
    }

}