/*

​

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom

​

​

                       1                         <------

                      / \

                     2   3                       <------

                      \   \

                      5    4                     <------

​

  expect output is 1, 3, 4

                        1                         <------

                      / 

                     2                          <------

                      \   

                      5                        <------

​

  expect output is 1, 2, 5

                       1                         <------

                      / \

                     2   3                       <------

                      \   

                      5                         <------

  expect output is 1, 3, 5

​

*/

using System;

using System.Collections.Generic;

​

public class Solution

{

    public class TreeNode

    {

        public int val;

        public TreeNode left;

        public TreeNode right;

        public TreeNode(int val = 0, TreeNode left = null, TreeNode right = null)

        {

            this.val = val;

            this.left = left;

            this.right = right;

        }

    }

​

    public static void Main(string[] args)

    {

        TreeNode root = new TreeNode(1);

        root.left = new TreeNode(2);

        root.right = new TreeNode(3);

        root.left.right = new TreeNode(5);

        root.right.right = new TreeNode(4);

        int[] result = RightSideView(root);

        if (result != null && result.Length > 0)

        {

            for (int i = 0; i < result.Length; i++)

            {

                System.Console.Out.WriteLine(result[i]);

            }

        }

    }

​

    public static int[] RightSideView(TreeNode root)

    {

        List<int> resultList = new List<int>();

        Queue<TreeNode> queue = new Queue<TreeNode>();

        TreeNode currentNode = root;

        queue.Enqueue(currentNode);

        while(queue.Count > 0){

            int n = queue.Count;

            for(int i=1; i<= n; i++){

                currentNode = queue.Dequeue();

            if(i == n){

                resultList.Add(currentNode.val);

            }

            if(currentNode.left != null)

                queue.Enqueue(currentNode.left);

            if(currentNode.right != null)

                queue.Enqueue(currentNode.right);

            }

        }

        return resultList.ToArray();

    }

}