Check mirror in n-ary tree

Difficulty Level : Hard
Last Updated : 10 Dec, 2020

Given two n-ary trees, the task is to check if they are mirror of each other or not. Print “Yes” if they are 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 2
Edge 2 of second N-ary: 1 3
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

The idea is to use Queue and Stack to check if given N-ary tree are mirror of each other or not.
Let first n-ary tree be t1 and second n-ary tree is t2. For each node in t1, make stack and push its connected node in it. Now, for each node in t2, make queue and push its connected node in it.
Now, for each corresponding node do following:

  While stack and Queue is not empty.
  a = top element of stack;
  b = front of queue;
  if (a != b)
    return false;
  pop element from stack and queue.

CPP

// C++ program to check if two n-ary trees are
// mirror.
#include <bits/stdc++.h>
using namespace std;
 
// First vector stores all nodes and adjacent of every 
// node in a stack.
// Second vector stores all nodes and adjacent of every 
// node in a queue.
bool mirrorUtil(vector<stack<int> >& tree1,
                      vector<queue<int> >& tree2)
{
    // Traversing each node in tree.
    for (int i = 1; i < tree1.size(); ++i) {
        stack<int>& s = tree1[i];
        queue<int>& q = tree2[i];
 
        // While stack is not empty && Queue is not empty
        while (!s.empty() && !q.empty()) {
 
            // checking top element of stack and front
            // of queue.
            if (s.top() != q.front())
                return false;
 
            s.pop();
            q.pop();
        }
 
        // If queue or stack is not empty, return false.
        if (!s.empty() || !q.empty())
            return false;
    }
 
    return true;
}
 
// Returns true if given two trees are mirrors.
// A tree is represented as two arrays to store
// all tree edges.
void areMirrors(int m, int n, int u1[], int v1[], 
                            int u2[], int v2[])
{
    vector<stack<int> > tree1(m + 1);
    vector<queue<int> > tree2(m + 1);
 
    // Pushing node in the stack of first tree.
    for (int i = 0; i < n; i++) 
        tree1[u1[i]].push(v1[i]);    
 
    // Pushing node in the queue of second tree.
    for (int i = 0; i < n; i++) 
        tree2[u2[i]].push(v2[i]);
     
 
    mirrorUtil(tree1, tree2) ? (cout << "Yes" << endl) :
                               (cout << "No" << endl);
}
 
// Driver code
int main()
{
    int M = 3, N = 2;
 
    int u1[] = { 1, 1 };
    int v1[] = { 2, 3 };
 
    int u2[] = { 1, 1 };
    int v2[] = { 3, 2 };
 
    areMirrors(M, N, u1, v1, u2, v2);
 
    return 0;
}

Python3

# Pytho3 program to check if two n-ary trees are
# mirror.
  
# First vector stores all nodes and adjacent of every 
# node in a stack.
# Second vector stores all nodes and adjacent of every 
# node in a queue.
def mirrorUtil(tree1, tree2):
 
    # Traversing each node in tree.
    for i in range(1, len(tree1)):
     
        s = tree1[i];
        q = tree2[i];
  
        # While stack is not empty && Queue is not empty
        while (len(s) != 0 and len(q) != 0):
  
            # checking top element of stack and front
            # of queue.
            if (s[-1] != q[0]):
                return False;
  
            s.pop();
            q.pop(0);
  
        # If queue or stack is not empty, return false.
        if(len(s) != 0 or len(q) != 0):
            return False;
  
    return True;
 
# Returns true if given two trees are mirrors.
# A tree is represented as two arrays to store
# all tree edges.
def areMirrors(m, n, u1, v1, u2, v2):
 
    tree1=[[] for i in range(m + 1)];
    tree2=[[] for i in range(m + 1)]
  
    # Pushing node in the stack of first tree.
    for i in range(n):
 
        tree1[u1[i]].append(v1[i]);    
  
    # Pushing node in the queue of second tree.
    for i in range(n):
     
        tree2[u2[i]].append(v2[i]);
      
    if mirrorUtil(tree1, tree2):
        print('Yes')
    else:
        print('No')
  
# Driver code
if __name__=='__main__':
 
    M = 3
    N = 2;
  
    u1 = [ 1, 1 ]
    v1 = [ 2, 3 ]
  
    u2 = [ 1, 1 ]
    v2 = [ 3, 2 ]
  
    areMirrors(M, N, u1, v1, u2, v2);
  
# This code is contributed by rutvik_56

Output:

Yes

Reference: https://practice.geeksforgeeks.org/problems/check-mirror-in-n-ary-tree/0
This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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