the binary tree for the above example will be
k(1)
\
a(2)
/ \
(7) a p(3)
\
i(4)
\
l(5)
\
r (6)
number in the bracket denotes the order of insertion like k inserted first
then a then p
and so on ..
now if inorder is performed ... arent we getting " kaapilr "
On Thu, Jun 23, 2011 at 12:43 AM, oppilas . <jatka.oppimi...@gmail.com>wrote:
> May be I didn't understood your logic.
> According to original question for
> I/P ----kapilra
> O/P --kaapilr..
> Now,
> -what if we create a binary tree with root as the first element of the
> string and if the next character is equal then place it to left else place
> it to right. Similar comparison will be done while inserting all the other
> nodes too .
>
> At root you will insert k(count=1)
> After that a
>
> k(1)
> /
> a(1)
> Then you will insert p,
> k(1)
> / \
> a(1) p(1)
> And so on.
>
>>
>>> after that if InOrder traversal is performed.. it would give us the
>>> desired output.
>>>
>>
> If by inoder traversal we can get desired output then please how me how
> using a small example :).
>
>
>>
>>> Snehi
>>>
>>>
>>> On Wed, Jun 22, 2011 at 9:48 PM, DK <divyekap...@gmail.com> wrote:
>>>
>>>> No. This is equivalent to a sort with comparisons based on index of
>>>> first occurrence in the input string. Any comparative algorithm is O(n log
>>>> n) and a non comparative algorithm can be O(n) only by using counting or
>>>> radix sorting etc.
>>>>
>>>> --
>>>> DK
>>>>
>>>> http://twitter.com/divyekapoor
>>>> http://www.divye.in
>>>>
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