driver(Tree *root)
{
node *array = malloc(sizeof(node *) *H ); /* take the height node
pointer array. this will act as source pointer of each list */
makeList(root, 0, array)
}
makeList(Tree *root, int level, node *array)
{
if(! root) return ;
/*do a Depth traversal and store the list pointer*/
append(array[level], root);
/*Traverse the left and right subtrees and store them in list*/
makeList(root->left, level+1, array);
makeList(root->right, level+1, array);
}
time complexity is O(n) space complexity O(log n)
( assuming append has O(1) complexity)
On Aug 1, 2:26 am, Nikhil Jindal <fundoon...@yahoo.co.in> wrote:
> @Ram Kumar: Yes. Simple and affective.
>
> Just at each node:
>
> node->left->side=node->right
> node->right->side=node->side->left
>
> i.e at each node you are setting the side of each of your child. Go on and
> just do it for all nodes. Done.
>
> On Sat, Jul 31, 2010 at 9:39 AM, Ram Kumar <harrypotter....@gmail.com>wrote:
>
>
>
>
>
> > DONT DO A BFS!! NOT WORTH IT! CALL THE NEW POINTER AS 'SIDE' POINTER.
>
> > for every root connect its left and right, for every root connect
> > root->right and root->SIDE->left
>
> > that ll do
>
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