# Check mirror in n-ary tree

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
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 stack;
if (a != b)
return false;
pop element from stack and queue.```

 `// C++ program to check if two n-ary trees are ` `// mirror. ` `#include ` `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 >& tree1, ` `                      ``vector >& 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 > tree1(m + 1); ` `    ``vector > 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; ` `} `

Output:

```Yes
```

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.