Check Mirror in N-ary tree.cpp

/*
Check Mirror in N-ary tree
===========================

Given two n-ary trees. Check if they are mirror images of each other or not. You are also given e denoting the number of edges in both trees, and two arrays, A[] and B[]. Each array has 2*e space separated values u,v denoting an edge from u to v for the both trees.

Example 1:
Input:
n = 3, e = 2
A[] = {1, 2, 1, 3}
B[] = {1, 3, 1, 2}
Output:
1
Explanation:
   1          1
 / \        /  \
2   3      3    2 
As we can clearly see, the second tree
is mirror image of the first.

Example 2:
Input:
n = 3, e = 2
A[] = {1, 2, 1, 3}
B[] = {1, 2, 1, 3}
Output:
1
Explanation:
   1          1
 / \        /  \
2   3      2    3 
As we can clearly see, the second tree
isn't mirror image of the first.

Your Task:
You don't need to read input or print anything. Your task is to complete the function checkMirrorTree() which takes 2 Integers n, and e;  and two arrays A[] and B[] of size 2*e as input and returns 1 if the trees are mirror images of each other and 0 if not.

Expected Time Complexity: O(n)
Expected Auxiliary Space: O(n)

Constraints:
1 <= n,e <= 105
*/

int checkMirrorTree(int n, int e, int A[], int B[])
{
  unordered_map<int, stack<int>> t1;
  for (int i = 0; i < 2 * e; i += 2)
  {
    int u = A[i], v = A[i + 1];
    t1[u].push(v);
  }

  for (int i = 0; i < 2 * e; i += 2)
  {
    int u = B[i], v = B[i + 1];
    if (t1[u].top() != v)
      return false;
    t1[u].pop();
  }
  return true;
}