Check mirror in n-ary tree

Given two n-ary trees, the task is to check if they are the mirror of each other or not. Print “Yes” if they are the mirror of each other else “No”.

Examples: 

Input : Node = 3, Edges = 2
Edge 1 of first N-ary: 1 2
Edge 2 of first N-ary: 1 3
Edge 1 of second N-ary: 1 3
Edge 2 of second N-ary: 1 2
Output : Yes

Input : Node = 3, Edges = 2
Edge 1 of first N-ary: 1 2 
Edge 2 of first N-ary: 1 3
Edge 1 of second N-ary: 1 2
Edge 2 of second N-ary: 1 3
Output : No

Approach  1: (Using Hashing)

The idea is to use an unordered map of stacks to check if given N-ary tree are mirror of each other or not. 
Let the first n-ary tree be t1 and the second n-ary tree is t2. For each node in t1, push its connected node in their corresponding stack in the map. Now, for each node in t2, their connected node match with the top of the stack, then pop elements from the stack.  

Otherwise, if the node does not match with the top of the stack then it means two trees are not mirror of each other. 

Now, for each corresponding node do the following:  

  1. Iterate over map of stack
      Push all connected nodes of each node of first tree in map of  stack.

  2. Again iterate over map for each node of second tree
      For example :     

      Let us take one node X of second tree 
      
      For this node X , check in map which stack is used

      a = Top of that stack for node X present in second tree;
      b = Connected node of X in second tree
      if (a != b)
           return false;
      pop node X from stack.

Implementation:

C++

// C++ program to check if two n-ary trees are
// mirror.
#include <bits/stdc++.h>
using namespace std;

// Function to check given two trees are mirror
// of each other or not
int checkMirrorTree(int M, int N, int u1[ ],
                    int v1[ ] , int u2[], int v2[])
    {
        // Map to store nodes of the tree
        unordered_map<int , stack<int>>mp;

        // Traverse first tree nodes
        for (int i = 0 ; i < N ; i++ )
        {
           mp[u1[i]].push(v1[i]);
        }

        // Traverse second tree nodes
        for (int i = 0 ; i < N ; i++)
        {
            if(mp[u2[i]].top() != v2[i])
                  return 0;
            mp[u2[i]].pop();
        }

        return 1;
    }

// Driver code
int main()
{
    int M = 7, N = 6;

    //Tree 1
    int u1[] = { 1, 1, 1, 10, 10, 10 };
    int v1[] = { 10, 7, 3, 4, 5, 6 };

    //Tree 2
    int u2[] = { 1, 1, 1, 10, 10, 10 };
    int v2[] = { 3, 7, 10, 6, 5, 4 };

    if(checkMirrorTree(M, N, u1, v1, u2, v2))
       cout<<"Yes";
    else
       cout<<"No";

    return 0;
}

Java

// Java program to check if two n-ary trees are mirror.
import java.util.*;
public class Main
{
    // Function to check given two trees are mirror
    // of each other or not
    static boolean checkMirrorTree(int M, int N, int[] u1, int[] v1, int[] u2, int[] v2)
    {

        // Map to store nodes of the tree
        HashMap<Integer, Stack<Integer>> mp = new HashMap<>();

        // Traverse first tree nodes
        for (int i = 0 ; i < N ; i++ )
        {
           if(!mp.containsKey(u1[i]))
           {
               mp.put(u1[i], new Stack<Integer>());
           }
           else{
               mp.get(u1[i]).push(v1[i]);
           }
        }

        // Traverse second tree nodes
        for (int i = 0 ; i < N ; i++)
        {
            if(mp.containsKey(u2[i]) && mp.get(u2[i]).size() > 0)
            {
                if(mp.get(u2[i]).peek() != v2[i])
                  return false;
                mp.get(u2[i]).pop();
            }
        }

        return true;
    }

  // Driver code
    public static void main(String[] args) {
        int M = 7, N = 6;

        // Tree 1
        int[] u1 = { 1, 1, 1, 10, 10, 10 };
        int[] v1 = { 10, 7, 3, 4, 5, 6 };

        // Tree 2
        int[] u2 = { 1, 1, 1, 10, 10, 10 };
        int[] v2 = { 3, 7, 10, 6, 5, 4 };

        if(checkMirrorTree(M, N, u1, v1, u2, v2))
           System.out.print("Yes");
        else
           System.out.print("No");
    }
}

// This code is contributed by divyeshrabadiya07.

Python3

# Python3 program to check if two n-ary trees are mirror.

# Function to check given two trees are mirror
# of each other or not
def checkMirrorTree(M, N, u1, v1, u2, v2):
    # Map to store nodes of the tree
    mp = {}

    # Traverse first tree nodes
    for i in range(N):
        if u1[i] in mp:
            mp[u1[i]].append(v1[i])
        else:
            mp[u1[i]] = []

    # Traverse second tree nodes
    for i in range(N):
        if u2[i] in mp and len(mp[u2[i]]) > 0:
            if(mp[u2[i]][-1] != v2[i]):
                return 0
            mp[u2[i]].pop()
    return 1

M, N = 7, 6

#Tree 1
u1 = [ 1, 1, 1, 10, 10, 10 ]
v1 = [ 10, 7, 3, 4, 5, 6 ]

#Tree 2
u2 = [ 1, 1, 1, 10, 10, 10 ]
v2 = [ 3, 7, 10, 6, 5, 4 ]

if(checkMirrorTree(M, N, u1, v1, u2, v2)):
   print("Yes")
else:
   print("No")

    # This code is contributed by rameshtravel07.

C#

// C# program to check if two n-ary trees are mirror.
using System;
using System.Collections.Generic;
class GFG {

    // Function to check given two trees are mirror
    // of each other or not
    static bool checkMirrorTree(int M, int N, int[] u1, int[] v1, int[] u2, int[] v2)
    {

        // Map to store nodes of the tree
        Dictionary<int, Stack<int>> mp = new Dictionary<int, Stack<int>>();

        // Traverse first tree nodes
        for (int i = 0 ; i < N ; i++ )
        {
           if(!mp.ContainsKey(u1[i]))
           {
               mp[u1[i]] = new Stack<int>();
           }
           else{
               mp[u1[i]].Push(v1[i]);
           }
        }

        // Traverse second tree nodes
        for (int i = 0 ; i < N ; i++)
        {
            if(mp.ContainsKey(u2[i]) && mp[u2[i]].Count > 0)
            {
                if(mp[u2[i]].Peek() != v2[i])
                  return false;
                mp[u2[i]].Pop();
            }
        }

        return true;
    }

  // Driver code
  static void Main()
  {
    int M = 7, N = 6;

    // Tree 1
    int[] u1 = { 1, 1, 1, 10, 10, 10 };
    int[] v1 = { 10, 7, 3, 4, 5, 6 };

    // Tree 2
    int[] u2 = { 1, 1, 1, 10, 10, 10 };
    int[] v2 = { 3, 7, 10, 6, 5, 4 };

    if(checkMirrorTree(M, N, u1, v1, u2, v2))
       Console.Write("Yes");
    else
       Console.Write("No");
  }
}

// This code is contributed by mukesh07.

Javascript

// JavaScript code for the above approach

// Function to check given two trees are mirror
// of each other or not
function checkMirrorTree(M, N, u1, v1, u2, v2) {
    // Map to store nodes of the tree
    let mp = {}

    // Traverse first tree nodes
    for (let i = 0; i < N; i++) {
        if (u1[i] in mp) {
            mp[u1[i]].push(v1[i]);
        } else {
            mp[u1[i]] = [v1[i]];
        }
    }

    // Traverse second tree nodes
    for (let i = 0; i < N; i++) {
        if (u2[i] in mp && mp[u2[i]].length > 0) {
            if (mp[u2[i]][mp[u2[i]].length - 1] != v2[i]) {
                return 0;
            }
            mp[u2[i]].pop();
        }
    }
    return 1;
}

let M = 7,
    N = 6;

//Tree 1
let u1 = [1, 1, 1, 10, 10, 10];
let v1 = [10, 7, 3, 4, 5, 6];

//Tree 2
let u2 = [1, 1, 1, 10, 10, 10];
let v2 = [3, 7, 10, 6, 5, 4];

if (checkMirrorTree(M, N, u1, v1, u2, v2)) {
    console.log("Yes");
} else {
    console.log("No");
}

// This code is contributed by Potta Lokesh

Output

Yes

Time Complexity: O(N).
Auxiliary  Space: O(N).

Approach 2: (Using LinkedList):

The main approach is to use one list of stack and one list of queue to store to value of nodes given in the form of two arrays.

1. Initialize both the lists with empty stack and empty queues respectively.
2. Now, iterate over the lists 
   Push all connected nodes of each node of first tree in list of stack and second tree list of queue.
3. Now iterate over the array and pop item from both stack and queue and check if they are same, if not same then return 0.

Implementation:

C++

// C++ program to check two n-ary trees are mirror.
#include <bits/stdc++.h>
using namespace std;

// Function to check given two trees are mirror
// of each other or not
int checkMirrorTree(int n, int e, int A[], int B[])
{
    //Lists to store nodes of the tree
    vector<stack<int>> s;
    vector<queue<int>> q;

    // initializing both list with empty stack and queue
    for (int i = 0; i <= n; i++)
    {
        s.push_back(stack<int>());
        queue<int> queue;
        q.push_back(queue);
    }

    // add all nodes of tree 1 to list of stack and tree 2 to list of queue
    for (int i = 0; i < 2 * e; i += 2)
    {
        s[A[i]].push(A[i + 1]);
        q[B[i]].push(B[i + 1]);
    }

    // now take out the stack and queues
    // for each of the nodes and compare them
    // one by one
    for (int i = 1; i <= n; i++)
    {
        while (!s[i].empty() && !q[i].empty())
        {
            int a = s[i].top();
            s[i].pop();
            int b = q[i].front();
            q[i].pop();

            if (a != b)
            {
                return 0;
            }
        }
    }

    return 1;
}

int main()
{
    int n = 3;
    int e = 2;
    int A[] = {1, 2, 1, 3};
    int B[] = {1, 3, 1, 2};

    if (checkMirrorTree(n, e, A, B) == 1)
    {
        cout << "Yes";
    }
    else
    {
        cout << "No";
    }

    return 0;
}
// This code is added by Srj_27

Java

// Java program to check two n-ary trees are mirror.

import java.io.*;
import java.util.*;

class GFG {

      // Function to check given two trees are mirror
    // of each other or not
      static int checkMirrorTree(int n, int e, int[] A, int[] B) {

          //Lists to store nodes of the tree
        List<Stack<Integer>> s = new ArrayList<>();
        List<Queue<Integer>> q = new ArrayList<>();

        // initializing both list with empty stack and queue
        for (int i = 0; i <= n; i++) {
            s.add(new Stack<>());
            Queue<Integer> queue = new LinkedList<>();
            q.add(queue);
        }

           // add all nodes of tree 1 to list of stack and tree 2 to list of queue
        for (int i = 0; i < 2 * e; i += 2) {
            s.get(A[i]).push(A[i + 1]);
            q.get(B[i]).add(B[i + 1]);
        }

          // now take out the stack and queues
        // for each of the nodes and compare them
        // one by one
        for (int i = 1; i <= n; i++) {
            while (!s.get(i).isEmpty() && !q.get(i).isEmpty()) {
                int a = s.get(i).pop();
                int b = q.get(i).poll();

                if (a != b) {
                    return 0;
                }
            }
        }

        return 1;
    }

    public static void main (String[] args) {
        int n = 3;
        int e = 2;
        int A[] = { 1, 2, 1, 3 };
        int B[] = { 1, 3, 1, 2 };

        if (checkMirrorTree(n, e, A, B) == 1) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }

    }
}

Python3

# Python3 program to check two n-ary trees are mirror.

# Function to check given two trees are mirror
# of each other or not
def checkMirrorTree(n, e, A, B):
    # Lists to store nodes of the tree
    s = []
    q = []

    # initializing both list with empty stack and queue
    for i in range(n + 1):
        s.append([])
        queue = []
        q.append(queue)

   # add all nodes of tree 1 to
   # list of stack and tree 2 to list of queue
    for i in range(0, 2 * e, 2):
        s[A[i]].append(A[i + 1])
        q[B[i]].append(B[i + 1])

    # now take out the stack and queues
    # for each of the nodes and compare them
    # one by one
    for i in range(1, n + 1):
        while (len(s[i]) > 0 and len(q[i]) > 0):
            a = s[i][len(s[i]) - 1]
            s[i].pop()
            b = q[i][0]
            q[i].pop(0)

            if (a != b):
                return 0
    return 1

  # Driver code
n = 3
e = 2
A = [ 1, 2, 1, 3 ]
B = [ 1, 3, 1, 2 ]

if (checkMirrorTree(n, e, A, B) == 1):
    print("Yes")
else:
    print("No")

    # This code is contributed by decode2207.

C#

// C# program to check two n-ary trees are mirror.
using System;
using System.Collections;
using System.Collections.Generic;
class GFG {

    // Function to check given two trees are mirror
    // of each other or not
    static int checkMirrorTree(int n, int e, int[] A, int[] B)
    {
        //Lists to store nodes of the tree
        List<Stack> s = new List<Stack>();
        List<Queue> q = new List<Queue>();

        // initializing both list with empty stack and queue
        for (int i = 0; i <= n; i++) {
            s.Add(new Stack());
            Queue queue = new Queue();
            q.Add(queue);
        }

           // add all nodes of tree 1 to list of stack and tree 2 to list of queue
        for (int i = 0; i < 2 * e; i += 2) {
            s[A[i]].Push(A[i + 1]);
            q[B[i]].Enqueue(B[i + 1]);
        }

          // now take out the stack and queues
        // for each of the nodes and compare them
        // one by one
        for (int i = 1; i <= n; i++) {
            while (s[i].Count > 0 && q[i].Count > 0) {
                int a = (int)s[i].Pop();
                int b = (int)q[i].Dequeue();

                if (a != b) {
                    return 0;
                }
            }
        }

        return 1;
    }

  static void Main() {
    int n = 3;
    int e = 2;
    int[] A = { 1, 2, 1, 3 };
    int[] B = { 1, 3, 1, 2 };

    if (checkMirrorTree(n, e, A, B) == 1) {
        Console.Write("Yes");
    } else {
        Console.Write("No");
    }
  }
}

// This code is contributed by divyesh072019.

Javascript

<script>
    // Javascript program to check two n-ary trees are mirror.

    // Function to check given two trees are mirror
    // of each other or not
      function checkMirrorTree(n, e, A, B) {

          //Lists to store nodes of the tree
        let s = [];
        let q = [];

        // initializing both list with empty stack and queue
        for (let i = 0; i <= n; i++) {
            s.push([]);
            let queue = [];
            q.push(queue);
        }

           // add all nodes of tree 1 to
           // list of stack and tree 2 to list of queue
        for (let i = 0; i < 2 * e; i += 2) {
            s[A[i]].push(A[i + 1]);
            q[B[i]].push(B[i + 1]);
        }

          // now take out the stack and queues
        // for each of the nodes and compare them
        // one by one
        for (let i = 1; i <= n; i++) {
            while (s[i].length > 0 && q[i].length > 0) {
                let a = s[i][s[i].length - 1];
                s[i].pop();
                let b = q[i][0];
                q[i].shift();

                if (a != b) {
                    return 0;
                }
            }
        }

        return 1;
    }

    let n = 3;
    let e = 2;
    let A = [ 1, 2, 1, 3 ];
    let B = [ 1, 3, 1, 2 ];

    if (checkMirrorTree(n, e, A, B) == 1) {
      document.write("Yes");
    } else {
      document.write("No");
    }

   // This code is contributed by suresh07.
</script>

Output

Yes

Time complexity: O(N) where N is no of nodes in given n-ary tree
Auxiliary space: O(N)

Reference: https://www.geeksforgeeks.org/problems/check-mirror-in-n-ary-tree/0

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