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;}
