Mirror a binary tree

 

Introduction

  Mirroring a binary tree, a common question nowadays to understand whether the candidate knows about tree traversal. Asking plainly in-order/pre-order is of no use!! :)

 

Solution 

The dummies way is to directly jump to code which never works even for such a simple problem.

First thought, i got this code:

Tree* mirror(Tree* node)

{

    if(node == NULL)

        return NULL;

    node->left = mirror(node->right);

    node->right = mirror(node->left);

    return node;

} 

Flaws in it?

- This is a mugged up code!!

- You didn’t think about whether to output the mirrored tree in a new data structure or to do it in-place

- Why you cannot use tail recursion for in-place,

- Reason 1: you need to swap left & right using a temporary storage. You need to expand the tree on the stack

- Reason 2: The above kind of swapping loses the node->left data whatever magic your recursion does.

 

Now, we got full of issues, think basic, non-iterative way

tmp = tree->left;

tree->left = tree->right;

tree->right = tmp;

Whatever way you follow, you need to do the above swap.

Corrected code for reproduction of mirrored tree, as a copy

Tree* mirror(Tree* node)

{

    if(node == NULL)

        return NULL;

    Tree *newNode = createNode(node->data);

    newNode->left = mirror(node->right);

    newNode->right = mirror(node->left);

    return newNode;

}

Correct code for creating mirrored tree, in-place

- do a post order traversal to read the full tree into the recursion stack

- do the primitive swap operation

Tree* mirror(Tree* node)

{

    if(node == NULL)

        return NULL;

    mirror(node->left);

    mirror(node->right);

    Tree *tmp = node->left;

    node->left = node->right;

    node->right = tmp;

    return node;

}