@Kaushik,

 I think you can do using inorder successor also, using this successor you
can think of BST as Sorted List.

 What do you say ?

Thanks,
Sathaiah


On Wed, May 19, 2010 at 11:11 AM, kaushik sur <kaushik....@gmail.com> wrote:

> @Dhilip
> Is it tested ? I doubt your code won't work ?
> @Rohit
> Can we anyways modify Morris Inorder Traversal process? We can have two
> pointers slow(increments once) and fast(increments twice), so that if fast
> reaches end or fast->next is end, we can have the median @ slow ?
>
> Correct me If I am wrong.
>
> Thanks and Regards
> Kaushik
>
>
>
> On Tue, May 18, 2010 at 4:05 PM, dhilip <dhilip.i...@gmail.com> wrote:
>
>> 1)do inorder and reverse inorder traversal
>> 2)They will meet at one point or they will cross each other
>> 3)That point is the median
>> 4)Code for the same.
>>
>> while(true)
>> {
>>  //inorder traversal
>>  while(count1<=count2 && flag1)
>>  {
>>   if(root)
>>   {
>>     push(root);
>>     root=root->lptr;
>>   }
>>   else
>>   {
>>     if(!isEmpty(stack1))
>>       t=pop();
>>     else
>>       flag1=false;
>>     var1=t->data;
>>     count1++;
>>     root=t->rptr;
>>   }
>>   if(count1==count2)
>>   {
>>     if(var1>=var2)
>>         return var2;
>>    }
>>
>>
>>  }
>>  //reverse inorder
>>  while(count2<=count1 && flag2)
>>  {
>>  if(root1)
>>   {
>>     push(root1);
>>     root1=root1->rptr;
>>   }
>>   else
>>   {
>>     if(!isEmpty(stack2))
>>       t1=pop();
>>     else
>>       flag2=false;
>>     var2=t1->data;
>>     count2++;
>>     root1=t1->lptr;
>>   }
>>    if(count1==count2)
>>    {
>>     if(var1>=var2)
>>      return var2;
>>
>>   }
>>  }
>>
>>
>> }
>>
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