Saturday, 4 May 2013

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;
}