**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;`

}

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