@vishal : as discussed in previous post , your solution wont work for
certain test cases...and i dont think so , checking tree in inorder way is
complex .It is simple to implement , you just need to call tree recursively
in Inorder way and  keep track of prev node and compare prev node with
current node if prev node < current then fine move ahead or report fail.

On Tue, Nov 6, 2012 at 12:39 PM, vishal chaudhary
<[email protected]>wrote:

> hi all,
> yes you can do it that way, but the thing is why are you increasing the
> complexity of the problem by again checking the inorder traversal output to
> be checked for increasing order.
> just traverse through the ones recursively(as we do it in the inoder
> traversal) through all the nodes and check whether the left child is less
> than the root and root is smaller than the right node.
>
>
> Warm Regards
> Vishal Chaudhary
> BE(Hons) Computer Science and Engineering
> BITS Pilani
>
>
>
>
>
>
> 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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