Check if two trees are Mirror | Set 2

Given two Binary Trees, returns true if two trees are mirror of each other, else false. 

Mirror Tree :

Previously discussed approach is here.

Approach: Find the inorder traversal of both the Binary Trees, and check whether one traversal is reverse of another or not. If they are reverse of each other then the trees are mirror of each other, else not. 

Implementation

Javascript

<script>

// JavaScript code to check two binary trees are
// mirror.

class Node
{
   constructor()
   {
     this.data = 0;
     this.left = null;
     this.right = null;
   }
};

// inorder traversal of Binary Tree
function inorder(n, v)
{
    if (n.left != null)
        inorder(n.left, v);

    v.push(n.data);

    if (n.right != null)
        inorder(n.right, v);
}

// Checking if binary tree is mirror
// of each other or not.
function areMirror(a, b)
{
    if (a == null && b == null)
        return true;
    if (a == null || b == null)
        return false;

    // Storing inorder traversals of both
    // the trees.
    var v1 = [];
    var v2 = [];

    inorder(a, v1);
    inorder(b, v2);

    if (v1.length != v2.length)
        return false;

    // Comparing the two arrays, if they
    // are reverse then return 1, else 0
    for(var i = 0, j = v2.length - 1; j >= 0;
            i++, j--)

    if (v1[i] != v2[j])
        return false;

    return true;
}

// Helper function to allocate a new node
function newNode(data)
{
    var node = new Node();
    node.data = data;
    node.left = node.right = null;
    return(node);
}

// Driver code
var a = newNode(1);
var b = newNode(1);

a.left = newNode(2);
a.right = newNode(3);
a.left.left  = newNode(4);
a.left.right = newNode(5);

b.left = newNode(3);
b.right = newNode(2);
b.right.left = newNode(5);
b.right.right = newNode(4);

if (areMirror(a, b))
{
    document.write("Yes");
}
else
{
    document.write("No");
}


</script>

Output

Yes

Time Complexity: O(NlogN)

Auxiliary Space: O(N)