@Ritu,
Right ! I misread you post

On Wed, Jan 26, 2011 at 3:44 PM, Ritu Garg <ritugarg.c...@gmail.com> wrote:

> @Algoose
>
> I said ..*.For every node x,go to the last node in its left subtree and
> mark the right child of that node as x.*
>
> it is to be repeated for all nodes except leaf nodes.
> to apply this approach ,you need to go down the tree.No parent pointers
> required.
> for every node say x whose left sub tree is not null ,go to the largest
> node in left sub-tree say y.
> Set  y->right = x
> y is the last node to be processed in left sub-tree of x hence x is
> successor of y.
>
> On Wed, Jan 26, 2011 at 3:27 PM, Algoose chase <harishp...@gmail.com>wrote:
>
>> @ritu
>> how would you find a successor without extra space if you dont have a
>> parent pointer ?
>> for Instance from the right most node of left subtree to the parent of
>> left subtree(root) ?
>> @Juver++
>> Internal stack does count as extra space !!
>>
>>
>> On Wed, Jan 26, 2011 at 3:02 PM, ritu <ritugarg.c...@gmail.com> wrote:
>>
>>> 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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