Binary Tree Maximum Path Sum

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
Solution:

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