@Mohit : in ur code u are breaking out when there parent reaches root. But
not every time will we have to reach root. For example:  Leftmost right
cousin of 1 is 9...in dat case parent should stop at 4. which is not the
root.
*Muthuraj R
IV th Year , ISE
PESIT , Bangalore*



On Wed, Aug 10, 2011 at 3:26 PM, muthu raj <[email protected]> wrote:

> @coder dumca: If u apply DFS  u will get a brother and not cousin. And yes
> it works only for some cases.
>
> *Muthuraj R
> IV th Year , ISE
> PESIT , Bangalore*
>
>
>
> On Wed, Aug 10, 2011 at 2:24 PM, coder dumca <[email protected]>wrote:
>
>> apply BFS
>> the node after the particular node(whode cousin to be find) will be the
>> required node
>>
>> On Wed, Aug 10, 2011 at 2:15 PM, Puneet Chawla <[email protected]
>> > wrote:
>>
>>>
>>> Agree with mohit goel..
>>>
>>> On Wed, Aug 10, 2011 at 11:22 AM, Mohit Goel 
>>> <[email protected]>wrote:
>>>
>>>>                        10
>>>>              4                      5
>>>>        2          7          6         11
>>>>    1     3    9   8     12  13   14   15
>>>>
>>>>
>>>> i think we should first  find the parent of the particular node ..then
>>>> apply the concept as told by Brijesh on it ....
>>>>
>>>> p =parent(q);
>>>> r = parent(p);
>>>> count =1;
>>>> while(p ==isright(r))
>>>> {
>>>> p=r;
>>>> r=parent(r);
>>>> count++;
>>>> if(r==root)
>>>> break;
>>>>
>>>> }
>>>>
>>>> if(d =right(r))
>>>> {
>>>> while(count!=0)
>>>> {
>>>> if(d->left)
>>>> d=d->left;
>>>> else d=d->right;
>>>> count--;
>>>> }
>>>> }
>>>> else return NULL;
>>>> o/p=d->value;
>>>>
>>>>
>>>>
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>>>
>>>
>>>
>>> --
>>> With regards
>>>   ............
>>> Puneet Chawla
>>> Computer Engineering Student
>>> NIT Kurukshetra
>>>
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