Mirror image of Binary tree

 Given a Binary Tree, convert it into its mirror.

            

Example 1:

Input:
      1
    /  \
   2    3
Output: 3 1 2
Explanation: The tree is
   1    (mirror)  1
 /  \    =>      /  \
2    3          3    2
The inorder of mirror is 3 1 2

Example 2:

Input:
      10
     /  \
    20   30
   /  \
  40  60
Output: 30 10 60 20 40
Explanation: The tree is
      10               10
    /    \  (mirror) /    \
   20    30    =>   30    20
  /  \                   /   \
 40  60                 60   40
The inroder traversal of mirror is
30 10 60 20 40.

Your Task:
Just complete the function mirror() that takes node as paramter  and convert it into its mirror. The printing is done by the driver code only.

Expected Time Complexity: O(N).
Expected Auxiliary Space: O(Height of the Tree).

Constraints:
1 ≤ Number of nodes ≤ 105
1 ≤ Data of a node ≤ 105

Ans :

void swapNode(Node *a , Node *b)

{

    Node *c ;

    c = a;

    a = b ;

    b = c;

}

class Solution {

  public:

    // Function to convert a binary tree into its mirror tree.

    void mirror(Node* node) {

        

        queue<Node *> q;

        q.push(node);

        while(!q.empty())

        {

            if(q.front()->left)

            q.push(q.front()->left);

            if(q.front()->right)

            q.push(q.front()->right);

            swap(q.front()->left , q.front()->right );

            q.pop();

        }

    }

};

Refrences : https://practice.geeksforgeeks.org/problems/mirror-tree/1#