@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 <[email protected]> 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 <[email protected]> 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 <[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 ? >>>>> >>>>> -- >>>>> You received this message because you are subscribed to the Google >>>>> Groups "Algorithm Geeks" group. >>>>> To post to this group, send email to [email protected]. >>>>> To unsubscribe from this group, send email to >>>>> [email protected]. >>>>> For more options, visit this group at >>>>> http://groups.google.com/group/algogeeks?hl=en. >>>>> >>>> >>>> >>>> >>>> -- >>>> best wishes!! >>>> Vaibhav >>>> >>>> -- >>>> You received this message because you are subscribed to the Google >>>> Groups "Algorithm Geeks" group. >>>> To post to this group, send email to [email protected]. >>>> To unsubscribe from this group, send email to >>>> [email protected]. >>>> For more options, visit this group at >>>> http://groups.google.com/group/algogeeks?hl=en. >>>> >>> >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Algorithm Geeks" group. >>> To post to this group, send email to [email protected]. >>> To unsubscribe from this group, send email to >>> [email protected]. >>> For more options, visit this group at >>> http://groups.google.com/group/algogeeks?hl=en. >>> >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to [email protected]. >> To unsubscribe from this group, send email to >> [email protected]. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- best wishes!! Vaibhav -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
