No,no extra space is needed.
Right children which are NULL pointers are replaced with pointer to
successor.

On Jan 26, 1:18 pm, nphard nphard <nphard.nph...@gmail.com> wrote:
> If you convert the given binary tree into right threaded binary tree, won't
> you be using extra space while doing so? Either the given tree should
> already be right-threaded (or with parent pointers at each node) or internal
> stack should be allowed for recursion but no extra space usage apart from
> that.
>
> On Wed, Jan 26, 2011 at 3:04 AM, ritu <ritugarg.c...@gmail.com> wrote:
> > it can be done in O(n) time using right threaded binary tree.
> > 1.Convert the tree to right threaded tree.
> > right threaded tree means every node points to its successor in
> > tree.if right child is not NULL,then it already contains a pointer to
> > its successor Else it needs to filled up as following
> >      a. For every node x,go to the last node in its left subtree and
> > mark the right child of that node as x.
> > It Can be done in O(n) time if tree is a balanced tree.
>
> > 2. Now,Traverse the tree with Inorder Traversal without using
> > additional space(as successor of any node is available O(1) time) and
> > keep track of 5th largest element.
>
> > Regards,
> > Ritu
>
> > On Jan 26, 8:38 am, nphard nphard <nphard.nph...@gmail.com> wrote:
> > > Theoretically, the internal stack used by recursive functions must be
> > > considered for space complexity.
>
> > > On Mon, Jan 24, 2011 at 5:40 AM, juver++ <avpostni...@gmail.com> wrote:
> > > > internal stack != extra space
>
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