it can easily be done with help of level order traversal.....

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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>
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-- 
**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD

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