^ 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 <[email protected]> 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 <[email protected]>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 <[email protected]> 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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>>
>>
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>>  Vaibhav
>>
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