Check if two binary trees are mirror

Given two arrays, A[] and B[] consisting of M pairs, representing the edges of the two binary trees of N distinct nodes according to the level order traversal, the task is to check if trees are the mirror images of each other.

Examples:

Input: N = 6, M = 5, A[][2] = {{1, 5}, {1, 4}, {5, 7}, {5, 8}, {4, 9}}, B[][2] = {{1, 4}, {1, 5}, {4, 9}, {5, 8}, {5, 7}}
Output: Yes
Explanation:

Example 1

Input: N = 5, M = 4, A[][2] = {{10, 20}, {10, 30}, {20, 40}, {20, 50}}, B[][2] = {{10, 30}, {10, 20}, {20, 40}, {20, 50}}
Output: No
Explanation:

Example 2

Set 1 and Set 2 of this article have been discussed in previous articles.

Approach: The given problem can be solved using Map and Set data structures. Follow the steps below to solve the problem:

Initialize two, maps of vectors say T1 and T2 to store the adjacency list of trees A and B respectively.
Initialize a set say St to store all the unique nodes' values.
Traverse the array A[] using the variable i, and performing the following steps:
- Push the value A[i][1] into the vector T1[A[i][0]] and then append the A[i][0] and A[i][1] to the set St.
As the edges are given according to the level order traversal, therefore in tree A for all nodes, the left child was inserted first and then the right child was inserted.
Traverse the array B[] in reverse order using the variable i and performing the following steps:
- Push the value B[i][1] into the vector T2[B[i][0]] and then append the B[i][0] and B[i][1] to the set St.
As array B[] is traversed in reverse, therefore in the tree B for all nodes, the right child was inserted first and then the left child was inserted.
Now iterate over the of set St and check if the vectors of children of the current node in tree A are not equal to the vectors of children of the current node in tree B, then print "No" as the answer and then return.
Finally, if none of the above cases satisfy, then print "Yes" as the answer.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check whether two binary
// trees are mirror image of each other
// or not
string checkMirrorTree(int N, int M,
                       int A[][2], int B[][2])
{
    // Stores the adjacency list
    // of tree A
    map<int, vector<int> > T1;

    // Stores the adjacency list
    // of tree B
    map<int, vector<int> > T2;

    // Stores all distinct nodes
    set<int> st;

    // Traverse the array A[]
    for (int i = 0; i < M; i++) {

        // Push A[i][1] in the
        // vector T1[A[i][0]]
        T1[A[i][0]].push_back(A[i][1]);

        // Insert A[i][0] in the
        // set st
        st.insert(A[i][0]);

        // Insert A[i][1] in the
        // set st
        st.insert(A[i][1]);
    }

    // Traverse the array B[] in
    // reverse
    for (int i = M - 1; i >= 0; i--) {

        // Push B[i][1] in the
        // vector T2[B[i][0]]
        T2[B[i][0]].push_back(B[i][1]);

        // Insert B[i][0] in the
        // set st
        st.insert(B[i][0]);

        // Insert B[i][0] in the
        // set st
        st.insert(B[i][1]);
    }

    // Iterate over the set st
    for (auto node : st) {

        // If vector T1[node] is
        // not equals to T2[node]
        if (T1[node] != T2[node])
            return "No";
    }

    // Return "Yes" as
    // the answer
    return "Yes";
}

// Driver Code
int main()
{
    // Given Input
    int N = 6;
    int M = 5;

    int A[][2] = {
        { 1, 5 }, { 1, 4 }, { 5, 7 }, { 5, 8 }, { 4, 9 }
    };
    int B[][2] = {
        { 1, 4 }, { 1, 5 }, { 4, 9 }, { 5, 8 }, { 5, 7 }
    };

    // Function Call
    cout << checkMirrorTree(N, M, A, B);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class Main
{
    // Function to check whether two binary
    // trees are mirror image of each other
    // or not
    static String checkMirrorTree(int N, int M, int[][] A, int[][] B)
    {
       
        // Stores the adjacency list
        // of tree A
        HashMap<Integer, Vector<Integer>> T1 = new HashMap<Integer, Vector<Integer>>();
      
        // Stores the adjacency list
        // of tree B
        HashMap<Integer, Vector<Integer>> T2 = new HashMap<Integer, Vector<Integer>>();
      
        // Stores all distinct nodes
        Set<Integer> st = new HashSet<Integer>();
      
        // Traverse the array A[]
        for (int i = 0; i < M; i++) {
      
            // Push A[i][1] in the
            // vector T1[A[i][0]]
            if(T1.containsKey(A[i][0]))
            {
                T1.get(A[i][0]).add(A[i][1]);
            }
            else{
                T1.put(A[i][0], new Vector<Integer>());
                T1.get(A[i][0]).add(A[i][1]);
            }
      
            // Insert A[i][0] in the
            // set st
            st.add(A[i][0]);
      
            // Insert A[i][1] in the
            // set st
            st.add(A[i][1]);
        }
      
        // Traverse the array B[] in
        // reverse
        for (int i = M - 1; i >= 0; i--) {
      
            // Push B[i][1] in the
            // vector T2[B[i][0]]
            if(T2.containsKey(B[i][0]))
            {
                T2.get(B[i][0]).add(B[i][1]);
            }
            else{
                T2.put(B[i][0], new Vector<Integer>());
                T2.get(B[i][0]).add(B[i][1]);
            }
      
            // Insert B[i][0] in the
            // set st
            st.add(B[i][0]);
      
            // Insert B[i][0] in the
            // set st
            st.add(B[i][1]);
        }
      
        // Iterate over the set st
        for(int node : st) {
      
            // If vector T1[node] is
            // not equals to T2[node]
            if (!(T1.get(node) == T2.get(node)))
                return "Yes";
        }
      
        // Return "No" as
        // the answer
        return "No";
    }
    
    public static void main(String[] args) {
        // Given Input
        int N = 6;
        int M = 5;
      
        int[][] A = {
            { 1, 5 }, { 1, 4 }, { 5, 7 }, { 5, 8 }, { 4, 9 }
        };
        int[][] B = {
            { 1, 4 }, { 1, 5 }, { 4, 9 }, { 5, 8 }, { 5, 7 }
        };
      
        // Function Call
        System.out.print(checkMirrorTree(N, M, A, B));
    }
}

// This code is contributed by rameshtravel07.

Python3

# Py program for the above approach

# Function to check whether two binary
# trees are mirror image of each other
# or not
def checkMirrorTree(N, M, A, B):
  
    # Stores the adjacency list
    # of tree A
    T1 = [[] for i in range(100)]

    # Stores the adjacency list
    # of tree B
    T2 = [[] for i in range(100)]

    # Stores all distinct nodes
    st = {}

    # Traverse the array A[]
    for i in range(M):
      
        # Push A[i][1] in the
        # vector T1[A[i][0]]
        T1[A[i][0]].append(A[i][1])

        # Insert A[i][0] in the
        # set st
        st[A[i][0]] = 1

        # Insert A[i][1] in the
        # set st
        st[A[i][1]] = 1
        
    # Traverse the array B[] in
    # reverse
    for i in range(M - 1, -1, -1):
        # Push B[i][1] in the
        # vector T2[B[i][0]]
        T2[B[i][0]].append(B[i][1])

        # Insert B[i][0] in the
        # set st
        st[B[i][0]] = 1

        # Insert B[i][0] in the
        # set st
        st[B[i][1]] = 1

    # Iterate over the set st
    for node in st:

        # If vector T1[node] is
        # not equals to T2[node]
        if (T1[node] != T2[node]):
            return "No"

    # Return "Yes" as
    # the answer
    return "Yes"

# Driver Code
if __name__ == '__main__':
    # Given Input
    N = 6
    M = 5

    A =[ [1, 5], [1, 4], [5, 7], [5, 8], [4, 9]]
    B =[ [ 1, 4 ],[ 1, 5 ],[ 4, 9 ],[ 5, 8 ],[ 5, 7 ]]

    # Function Call
    print (checkMirrorTree(N, M, A, B))

    # This code is contributed by mohit kumar 29.

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
    
    // Function to check whether two binary
    // trees are mirror image of each other
    // or not
    static string checkMirrorTree(int N, int M,
                           int[,] A, int[,] B)
    {
      
        // Stores the adjacency list
        // of tree A
        Dictionary<int, List<int>> T1 = new Dictionary<int, List<int>>();
     
        // Stores the adjacency list
        // of tree B
        Dictionary<int, List<int>> T2 = new Dictionary<int, List<int>>();
     
        // Stores all distinct nodes
        HashSet<int> st = new HashSet<int>();
     
        // Traverse the array A[]
        for (int i = 0; i < M; i++) {
     
            // Push A[i][1] in the
            // vector T1[A[i][0]]
            if(T1.ContainsKey(A[i,0]))
            {
                T1[A[i,0]].Add(A[i,1]);
            }
            else{
                T1[A[i,0]] = new List<int>();
                T1[A[i,0]].Add(A[i,1]);
            }
     
            // Insert A[i][0] in the
            // set st
            st.Add(A[i,0]);
     
            // Insert A[i][1] in the
            // set st
            st.Add(A[i,1]);
        }
     
        // Traverse the array B[] in
        // reverse
        for (int i = M - 1; i >= 0; i--) {
     
            // Push B[i][1] in the
            // vector T2[B[i][0]]
            if(T2.ContainsKey(B[i,0]))
            {
                T2[B[i,0]].Add(B[i,1]);
            }
            else{
                T2[B[i,0]] = new List<int>();
                T2[B[i,0]].Add(B[i,1]);
            }
     
            // Insert B[i][0] in the
            // set st
            st.Add(B[i,0]);
     
            // Insert B[i][0] in the
            // set st
            st.Add(B[i,1]);
        }
     
        // Iterate over the set st
        foreach(int node in st) {
     
            // If vector T1[node] is
            // not equals to T2[node]
            if (!T1[node].Equals(T2[node]))
                return "Yes";
        }
     
        // Return "No" as
        // the answer
        return "No";
    }

  static void Main()
  {
    
    // Given Input
    int N = 6;
    int M = 5;
 
    int[,] A = {
        { 1, 5 }, { 1, 4 }, { 5, 7 }, { 5, 8 }, { 4, 9 }
    };
    int[,] B = {
        { 1, 4 }, { 1, 5 }, { 4, 9 }, { 5, 8 }, { 5, 7 }
    };
 
    // Function Call
    Console.Write(checkMirrorTree(N, M, A, B));
  }
}

// This code is contributed by divyesh072019.

JavaScript

<script>
    // Javascript program for the above approach
    
    // Function to check whether two binary
    // trees are mirror image of each other
    // or not
    function checkMirrorTree(N, M, A, B)
    {
    
        // Stores the adjacency list
        // of tree A
        let T1 = []; 

        // Stores the adjacency list
        // of tree B
        let T2 = [];
        
        for(let i = 0; i < 100; i++)
        {
            T1.push([]);
            T2.push([]);
        }

        // Stores all distinct nodes
        let st = new Map();

        // Traverse the array A[]
        for(let i = 0; i < M; i++)
        {
            // Push A[i][1] in the
            // vector T1[A[i][0]]
            T1[A[i][0]].push(A[i][1]);

            // Insert A[i][0] in the
            // set st
            st[A[i][0]] = 1;

            // Insert A[i][1] in the
            // set st
            st[A[i][1]] = 1;
        }

        // Traverse the array B[] in
        // reverse
        for(let i = M - 1; i < -1; i=-1)
        {
            // Push B[i][1] in the
            // vector T2[B[i][0]]
            T2[B[i][0]].push(B[i][1]);

            // Insert B[i][0] in the
            // set st
            st[B[i][0]] = 1;

            // Insert B[i][0] in the
            // set st
            st[B[i][1]] = 1;
        }

        // Iterate over the set st
        st.forEach((values,node)=>{
          // If vector T1[node] is
          // not equals to T2[node]
          if (T1[node] != T2[node])
          {
            return "No";
          }
        })

        // Return "Yes" as
        // the answer
        return "Yes";
    }
    
    // Given Input
    let N = 6;
    let M = 5;
 
    let A = [ [1, 5], [1, 4], [5, 7], [5, 8], [4, 9]];
    let B = [ [ 1, 4 ],[ 1, 5 ],[ 4, 9 ],[ 5, 8 ],[ 5, 7 ]];
 
    // Function Call
    document.write(checkMirrorTree(N, M, A, B));
    
    // This code is contributed by divyeshrabadiya07.
</script>

Output

Yes

Time Complexity: O((N+M)*log(N))
Auxiliary Space: O(N+M)

Check if two binary trees are mirror | Set 3

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