@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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