you can check if the element to be inserted next is same as the node value
while inserting itself. Why to do a separate inorder traversal..

On Fri, Jun 3, 2011 at 12:56 AM, Supraja Jayakumar <suprajasank...@gmail.com
> wrote:

> Hi
>
> Inserting into BST and break on same element will detect duplicates.
> But in any case, this is still O(2n) ? - once for inserting them in to the
> tree
> and another for the inorder read.
>
> Correct me if wrong.
>
> Rgds
> Supraja J
>
>
> On Thu, Jun 2, 2011 at 1:11 PM, Harshal <hc4...@gmail.com> wrote:
>
>> @Anurag: XOR wont work here, 1 element is repeated, not 1 element is
>> unique. Read the question again.
>>
>> Keep inserting elements in a BST and break once you find the same element.
>> O(nlogn)
>>
>>
>> On Fri, Jun 3, 2011 at 12:38 AM, Anurag Narain <anuragnar...@gmail.com>wrote:
>>
>>> take X-OR of all the elements.....the one which has no duplicate will be
>>> left and rest all will be reduced to zero.
>>>
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>>
>>
>> --
>> Harshal Choudhary,
>> III Year B.Tech CSE,
>> NIT Surathkal, Karnataka, India.
>>
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-- 
Harshal Choudhary,
III Year B.Tech CSE,
NIT Surathkal, Karnataka, India.

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