Given a non-empty binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

Example 1:

Input: [1,2,3]

       1
      / \
     2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxPathSum(TreeNode root) {
        if (root == null) {
            return Integer.MIN_VALUE;
        }
        
        int[] maxSum = { Integer.MIN_VALUE };
        maxPathSumHelper (root, maxSum);
        
        return maxSum [0];
    }
    
    private int maxPathSumHelper (TreeNode root, int[] maxSum) {
        if (root == null) {
            return 0;
        }
        
        // 如果得到的max是负数，那么对于求maxSum没有贡献，直接忽略，取成0
        int leftMax = Math.max (maxPathSumHelper (root.left, maxSum), 0);
        int rightMax = Math.max (maxPathSumHelper (root.right, maxSum), 0);
        int currentMax = root.val + leftMax + rightMax;
        
        // update maxSum
        maxSum [0] = Math.max (maxSum[0], currentMax);
        
        // return a path containning root node and make max path sum
        return root.val + Math.max (leftMax, rightMax); 
    }
}