We will use stack and queue data structure since stack follow LIFO that is last in first out the way and queue follow FIFO first in first out pattern, is the trees are mirror then the top of the stack will be equal to the front of the queue and if they aren't equal it means that they are not the mirror of each other.

We will follow this approach, taking each node at a time and checking its connected component in stack and queue. For checking whether each subtree in itself is a mirror or not we will use a boolean variable flag, initially, the flag is true and each time we check if the top of stack and queue front are equal or not, if not then simply return NO as the answer and after checking all nodes return true if all are valid nodes.

C++ Implementation:

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

int main()
{
    ll t;

    cout << "Enter number of test cases: ";
    cin >> t;

    while (t--) {
        ll n, m;

        cout << "Enter number of nodes and edges: ";
        cin >> n >> m;

        // declare a vector of stacks of size (n+1)
        // since index start from 0.
        stack<int> v1[n + 1];
        // declare a vector of queue of size (n+1)
        // since index start from 0.
        queue<int> v2[n + 1];

        cout << "Enter tree 1: ";
        for (ll i = 0; i < m; i++) {
            ll x, y; //enter the elements of its.
            cin >> x >> y;
            v1[x].push(y);
        }

        cout << "Enter tree 2: ";
        for (ll i = 0; i < m; i++) {
            ll x, y;
            cin >> x >> y; //enter elements of tree 2.
            v2[x].push(y);
        }
        bool flag;
        // iterate through each node.
        for (int i = 1; i <= n; i++) {
            // store nodes of tree 1 connected components in stack
            stack<int>& s = v1[i];
            // store nodes of tree 1 connected components in queue
            queue<int>& q = v2[i];

            // declare a boolean variable to check mirror
            // property among the elements.
            flag = true;
            //compare the stack top to queue top.
            while (!s.empty() and !q.empty()) {
                // if not similar then break from the loop.
                if (s.top() != q.front()) {
                    flag = false;
                    break;
                }
                s.pop();
                q.pop();
            }
            // if not similar then break from the nodes loop
            // since no further comparison is needed.
            if (flag == false)
                break;
        }

        // check if mirror or not.
        cout << "Is Mirror: ";
        if (flag == true)
            cout << "YES"
                 << "\n";
        else
            cout << "NO"
                 << "\n";
    }

    return 0;
}

Output:

Enter number of test cases: 3
Enter number of nodes and edges: 4 4 
Enter tree 1: 1 2 1 3 2 4 2 5
Enter tree 2: 1 2 1 3 2 5 2 4
Is Mirror: NO
Enter number of nodes and edges: 3 2
Enter tree 1: 1 2 1 3
Enter tree 2: 1 3 1 2
Is Mirror: YES
Enter number of nodes and edges: 3 3
Enter tree 1: 1 2 1 3 2 4
Enter tree 2: 1 3 1 2 2 4
Is Mirror: YES

• The time complexity for the above approach in worst case: O(n*n)
• Space complexity for the above approach in worst case : O(n)