Check this http://ideone.com/ARsJ1

On Thu, Jun 30, 2011 at 7:48 PM, oppilas . <jatka.oppimi...@gmail.com>wrote:

> I am unable to find a test case where this particular approach fails( I
> hardly thinks it's correct but anyway here it is).
>
> We make the last element the root of the BST.
> And keep inserting each element one by one  just like normal binary tree
> insertion.
> This will take O(n^2) time in worst case( when the tree is only leftist or
> rightist). For a balance tree it will be T(nLogn).
>
>
> On Thu, Jun 30, 2011 at 7:22 PM, sunny agrawal <sunny816.i...@gmail.com>wrote:
>
>> it can be done without sorting(Finding any other traversal)
>>
>> Do it recursively,
>> last element of the traversal will be head, and now if you will see in the
>> traversal left part of the traversal will be its LST and Right will be RST
>> (except Head) only thing you need to find will be the index of the element
>> which divides the traversal into LST and RST. this index can be find very
>> easily beacause all elements of LST are less than root value.
>>
>>
>> On Thu, Jun 30, 2011 at 7:16 PM, Nishant <mittal.nishan...@gmail.com>wrote:
>>
>>> if BST contains integers then sort the postorder traversal which will
>>> give you inorder traversal...
>>>
>>> On Jun 30, 6:27 pm, "oppilas ." <jatka.oppimi...@gmail.com> wrote:
>>> > Is it possible to create a binary search tree (not binary tree) from
>>> post
>>> > order traversal only?
>>> > If give, how and if not please give reason/counter examples.
>>>
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>>
>>
>> --
>> Sunny Aggrawal
>> B-Tech IV year,CSI
>> Indian Institute Of Technology,Roorkee
>>
>>
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