@atul anand :  actually i was storing the inorder in a seperate array and
then was checking whether it was in inceasing order(like the very first
post in this thread).. hence extra space...
To avoid that, i tried another approach of maitaining a prev pointer and
comparing it with current node and giving result based on that.


On Fri, Nov 9, 2012 at 10:00 AM, atul anand <atul.87fri...@gmail.com> wrote:

> @saurabh : correct..yes if you are considering recursive approach , so it
> will take O(n) space stack.But same can be done using Morris traversal then
> space will be constant.
>
> On Fri, Nov 9, 2012 at 7:40 AM, saurabh singh <saurab...@gmail.com> wrote:
>
>> ^ To perform inorder traversal in  a binary tree without using stack
>> space the tree must be mutable. In other cases as far as I can think the
>> space complexity should be asymptotically O(n) where n are the number of
>> nodes.
>>
>> Saurabh Singh
>> B.Tech (Computer Science)
>> MNNIT
>> blog:geekinessthecoolway.blogspot.com
>>
>>
>>
>> On Wed, Nov 7, 2012 at 10:09 AM, atul anand <atul.87fri...@gmail.com>wrote:
>>
>>> @vaibhav : by not using extra space...i guess you mean that you were not
>>> allowed to use one extra pointer.bcozz space complexity will remain
>>> constant for inorder approch.
>>>
>>> On Tue, Nov 6, 2012 at 1:07 AM, vaibhav shukla 
>>> <vaibhav200...@gmail.com>wrote:
>>>
>>>> yes ofcourse... dats the easiest i suppose...
>>>> but in one of my interviews, i told this approach, but was then asked
>>>> not to use space (which i was ,to store inorder)
>>>> So for such cases, you must try other approaches as well. (DO
>>>> inorder,keep track of previously visited node and compare it with current
>>>> node for value greater,or less accordingly.)
>>>>
>>>>
>>>> On Tue, Nov 6, 2012 at 12:34 AM, shady <sinv...@gmail.com> wrote:
>>>>
>>>>> Hi,
>>>>> Can we check this by just doing an inorder traversal, and then
>>>>> checking if it is in increasing order or not ?
>>>>>
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>>>>
>>>>
>>>> --
>>>> best wishes!!
>>>>  Vaibhav
>>>>
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best wishes!!
 Vaibhav

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