@jalaj: Use the same algorithm as finding the height of a normal tree (I
hope you're familiar with it) . But to know that you've reached a leaf, you
would normally use node->left or node->right ==NULL. But here in this case,
the leaves form a doubly linked list, so the former condition can no longer
be used and hence the condition becomes node==node->right->left which is
nothing but node==node->next->previous in doubly linked list terminology.

I hope this is clear.

On Sun, Jun 27, 2010 at 12:48 AM, Raj N <[email protected]> wrote:

> @Algoose: Thanks for correcting it. Even I have made the same mistake as
> Divya. It shud be node == node->right->left
>
>
> On Sat, Jun 26, 2010 at 7:20 PM, Algoose Chase <[email protected]>wrote:
>
>> If the parent Pointer is available we can use
>>
>> if( node->right->data > node->parent->data || node->right->data <
>> node->data)  to identify the leaf nodes and use it as one of the recursion
>> terminating condition in the original Height determining Algorithm.
>>
>> @Divya, I think u meant node != node->right->left
>>
>>
>> On Sat, Jun 26, 2010 at 3:54 PM, divya jain <[email protected]>wrote:
>>
>>> yes..
>>>
>>> i got the solution traverse till node->right!=node->right->left... at
>>> this point u ll get height.. rite?
>>>
>>>
>>> On 26 June 2010 11:49, Raj N <[email protected]> wrote:
>>>
>>>> @Divya: What will happen when say node->right when you reach the leaves
>>>> ? Is it equivalent to node->next and node->left = = node->previous in the
>>>> doubly linked list ?
>>>>
>>>>
>>>> On Tue, Jun 22, 2010 at 4:44 PM, divya <[email protected]>wrote:
>>>>
>>>>> a bst is given whose leaf nodes having left as well as right pointers
>>>>> not pointing to NULL. rather all the leaf nodes are forming a circular
>>>>> doubly linked list. u have to calculate height of tree.
>>>>>
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