Leetcode 114 - Flatten binary tree to list - a tough ride

Leetcode114_FlattenBinaryTreeToList_Practice.cs

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

namespace Leetcode114_FlattenBinaryTreeToLinkedList
{
    /// <summary>
    /// Leetcode 114 - Flatten binary tree to linked list
    /// 
    /// </summary>
    class Program
    {
        static void Main(string[] args)
        {
        }        
        
        //Definition for a binary tree node.
        public class TreeNode {
          public int val;
          public TreeNode left;
          public TreeNode right;
          public TreeNode(int x) { val = x; }
        }
 
        public class Solution
        {
            public static void Flatten(TreeNode root)
            {
                if (root == null)
                {
                    return; 
                }

                bool hasLeftChild  = root.left != null;
                bool hasRightChild = root.right != null; 
                if (hasLeftChild)
                {
                    var left = root.left;                   
           
                    Flatten(left);                                                           
                }

                if (hasRightChild)
                {
                    var right = root.right; // added after failed online judge
                    Flatten(right);

                    // set left child to null
                    right.left = null; 
                }

                if (hasLeftChild)
                {
                    var left = root.left;
                    var end  = getEnd(left);                    

                    var tmp = root.right;

                    root.right = left;
                    end.right  = tmp;

                    // set left child to null
                    root.left = null; // set left child to null  
                }                
            }

            /// <summary>
            /// 
            /// </summary>
            /// <param name="root"></param>
            public static TreeNode getEnd(TreeNode root)
            {
                bool hasRightChild = root.right != null;
                if (hasRightChild)
                {
                    return getEnd(root.right);
                }
                else
                {
                    return root; 
                }
            }
        }
    }
}

A few mistakes in the first writing to pass online judge.

1. Forget to flatten root.right children tree.
2. It takes me more than 30 minutes to understand that the left child should set to null after it is flattened.
3. Run into null pointer in the execution, so delay the root.left = null until line 65