I know dat, but i think inorder traversal solutn is better, cos in level
order case, it wud be difficult to find the subset tht is the largest BST..


@Siddharam : i think your solution is correct..

On Fri, Aug 5, 2011 at 10:20 AM, Dipankar Patro <dip10c...@gmail.com> wrote:

>
> checking for BST node is quite simple. by definition, p is a node of bst if
> (p->left < p < p->right)
> each node should satisfy this condition in BST. (except for the exterior
> nodes)
>
>
> On 5 August 2011 10:13, Aman Goyal <aman.goya...@gmail.com> wrote:
>
>> @siddharam: nice approach, just a query.... increasing inorder traversal
>> of a tree is a sufficient condition to check for a BST ?
>>
>>
>> On Fri, Aug 5, 2011 at 10:05 AM, siddharam suresh <
>> siddharam....@gmail.com> wrote:
>>
>>> my idea is "go for inorder traversal find the longest sorted sequence in
>>> traversal thats the *'largest BST in a binary tree.'* "
>>> Thank you,
>>> Siddharam
>>>
>>>
>>>
>>> On Fri, Aug 5, 2011 at 10:00 AM, Aman Goyal <aman.goya...@gmail.com>wrote:
>>>
>>>> while dequing  a node from the queue, how will u check whether a bst
>>>> property is sattisfied or not ?..
>>>>
>>>>
>>>> On Fri, Aug 5, 2011 at 9:49 AM, Dipankar Patro <dip10c...@gmail.com>wrote:
>>>>
>>>>> I have some upto this much currently.
>>>>> Modify the Breadth First traversal (BFT) a bit. maintain two queues,
>>>>> one is for original traversal.
>>>>>
>>>>> Start from root,  BFT. when you dequeue a node, check if it satisfies
>>>>> the condition for BST. if yes add the the node to auxiliary queue, if not,
>>>>> leave it and add it's original children to the original queue in both 
>>>>> cases.
>>>>> Some further modifications can the done to have multiple auxiliary
>>>>> queues and keep track of their heights.
>>>>>
>>>>> What say?
>>>>>
>>>>>
>>>>> On 5 August 2011 09:40, Aman Goyal <aman.goya...@gmail.com> wrote:
>>>>>
>>>>>> Yes, that can be a liable case definitely....!!!
>>>>>>
>>>>>>
>>>>>> On Fri, Aug 5, 2011 at 9:35 AM, Dipankar Patro 
>>>>>> <dip10c...@gmail.com>wrote:
>>>>>>
>>>>>>> The question is a bit tricky.
>>>>>>> Is it possible that the largest BST is somewhere in deeper depth,
>>>>>>> i.e. it is not necessarily consisting of the root?
>>>>>>>
>>>>>>> On 5 August 2011 08:46, Aman Goyal <aman.goya...@gmail.com> wrote:
>>>>>>>
>>>>>>>> How to find the largest BST in a binary tree.
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>>> 15
>>>>>>>> / \
>>>>>>>> 10__________ 20
>>>>>>>> / \
>>>>>>>> 5 _____7____
>>>>>>>> / \
>>>>>>>> 2_ __5
>>>>>>>> / \    /
>>>>>>>> 0 8 3
>>>>>>>>
>>>>>>>> The largest BST (may or may not include all of its descendants) from
>>>>>>>> the above example should be:
>>>>>>>>
>>>>>>>> ____15____
>>>>>>>> / \
>>>>>>>> _10 20
>>>>>>>> /
>>>>>>>> 5
>>>>>>>>
>>>>>>>>
>>>>>>>> Please do not post working code, logic/algorithm or link would be
>>>>>>>> preferred.
>>>>>>>> I know it will be through recursion  , still the logic part of
>>>>>>>> recursion is not clear.. would be thankful if anyone could help.
>>>>>>>>
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>>>>>>>
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