using System;

using System.Linq;

using System.Collections.Generic;

using System.Text;

​

/*

​

**************** This is optimized Solution *****************

​

Link: [https://www.scaler.com/academy/mentee-dashboard/class/26775/assignment/problems/228?navref=cl_tt_nv]

Problem Description

​

Given a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

​

Problem Constraints

1 <= number of nodes <= 105

​

Input Format

First and only argument is the root node of the binary tree.

​

Output Format

Return 0 / 1 ( 0 for false, 1 for true ).

​

Example Input

Input 1:

​

    1

   / \

  2   2

 / \ / \

3  4 4  3

Input 2:

​

    1

   / \

  2   2

   \   \

   3    3

​

​

Example Output

Output 1:

​

 1

Output 2:

​

 0

​

Example Explanation

Explanation 1:

​

 The above binary tree is symmetric. 

Explanation 2:

​

The above binary tree is not symmetric.

*/

​

public class TreeNode {

    public int val;

    public TreeNode left;

    public TreeNode right;

    public TreeNode(int x) { this.val = x; this.left = this.right = null; }

}

​

public class Program

{

    private static bool isMirrorTree(TreeNode A, TreeNode B)

    {

        if(A == null && B == null){

            return true;

        }

        if(A == null || B == null){

            return false;

        }

        return Program.isMirrorTree(A.left, B.right) && Program.isMirrorTree(A.right, B.left);

    }

    public static int isSymmetricTree(TreeNode A){

        bool isMirrorTree = Program.isMirrorTree(A.left, A.right); 

        if(isMirrorTree) { return 1; }else return 0;

    }

    public static void Main()

    {

        TreeNode A = new TreeNode(1);

        A.left = new TreeNode(2);

        A.right = new TreeNode(2);

        A.left.left = new TreeNode(3);

        A.left.right = new TreeNode(4);

        A.right.left = new TreeNode(3);

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

        /*TreeNode A = new TreeNode(5);

        A.left = new TreeNode(5);

        A.right = new TreeNode(10);

        TreeNode B = new TreeNode(3);

        B.left = new TreeNode(5);*/

        /*TreeNode A = new TreeNode(1);

        A.left = new TreeNode(2);

        A.right = new TreeNode(3);

        A.right.left = new TreeNode(4);

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

        TreeNode B = new TreeNode(1);

        B.left = new TreeNode(2);

        B.right = new TreeNode(3);

        B.right.left = new TreeNode(4);

        B.right.left.right = new TreeNode(7);*/

        int _isIdentical = Program.isSymmetricTree(A);

        Console.WriteLine("Is Binary Tree Identical: " + _isIdentical);

    }

}