Binary Tree Maximum Value end by root node - maximum value from root node to any node in the binary tree

BinaryTreeMaxValueEndByRoot1.cs

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace _BinaryTreeMaximumValueEndByRoot
{
    public class TreeNode
    {
        public int val;
        public TreeNode left;
        public TreeNode right;
        public TreeNode(int x) { val = x; }
    }

    class Program
    {
        static void Main(string[] args)
        {
            int result = testCase_1();
            int result2 = testCase_2();
            int result3 = testCase_3();
        }

        /*
         * maximum value is ended by root value
         */
        public static int testCase_1()
        {
            TreeNode n1 = new TreeNode(3);
            n1.left = new TreeNode(2);
            n1.left.left = new TreeNode(1);
            n1.left.right = new TreeNode(3);

            n1.right = new TreeNode(-1);
            n1.right.left = new TreeNode(-2);
            n1.right.right = new TreeNode(-2);

            return getMaximumEndByRoot(n1);
        }

        /*
         * maximum value is in subtree somewhere
         */
        public static int testCase_2()
        {
            TreeNode n1 = new TreeNode(-3);
            n1.left = new TreeNode(2);
            n1.left.left = new TreeNode(1);
            n1.left.right = new TreeNode(3);

            n1.right = new TreeNode(-1);
            n1.right.left = new TreeNode(-2);
            n1.right.right = new TreeNode(-2);

            return getMaximumEndByRoot(n1);
        }

        /*
         *  Test case: 
         *  9 6 -3 null null -6 2 null null -6 2 null null 2 null -6 -6 null null null null
         *  
         * Maximum path is 6 9 -3 2 2 
         */
        public static int testCase_3()
        {
            TreeNode n1 = new TreeNode(9);
            n1.left = new TreeNode(6);

            n1.right = new TreeNode(-3);
            n1.right.left = new TreeNode(-6);
            n1.right.right = new TreeNode(2);
            n1.right.right.left = new TreeNode(2);
            n1.right.right.left.left = new TreeNode(-6);
            n1.right.right.left.right = new TreeNode(-6);

            return getMaximumEndByRoot(n1);
        }

        /*
         * Analyze the binary tree maximum value ended by root node problem.    
         * 
         * Add some analysis and reasoning here:  
         * The problem is to solve the path from any node of the tree to another node of the tree.     
         * 
         * 
         */
        private static int getMaximumEndByRoot(TreeNode root )
        {
            if (root == null)
                return 0;

            int left  = getMaximumEndByRoot(root.left);
            int right = getMaximumEndByRoot(root.right);           

            // 3 values, get maximum one.          
            return Math.Max(root.val, Math.Max(root.val + left, root.val + right));
        }
    }
}