make a function which returns the missle of the LL
middle(node *)
now,
node * listtotree(node *head){
if( !head || !head->next){
return head;
node *temp= middle(head);
node *p=head;
while(p->next!=temp){
p=p->next;
}
p->next=NULL;
node *q=temp->next;
temp->next=NULL;
temp->prev=listtotree(head);
temp->next=listtotree(q);
return temp;
}
i think this will do :)
On Sun, Aug 1, 2010 at 10:57 AM, Shiv ... <shivsinha2...@gmail.com> wrote:
> No it won't it will just reduce the height of tree to n/2 (from n).
>
> My algo-
> 1. Parse in triplets. For every 3 nodes make the second one parent of other
> two. Also, queue up such parents.
> 2. repeat step 1 till you have only 1 node left (root).
>
> But this may give a tree 'imbalanced at root. we may need to do some height
> re-balancing.
>
> -Thanks
>
>
>
> On Sun, Aug 1, 2010 at 9:26 AM, Manjunath Manohar <
> manjunath.n...@gmail.com> wrote:
>
>> find the middle of the list and make it as the root..thus i this maner u
>> will get a balanced tree..
>>
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With Regards,
Jalaj Jaiswal
+919026283397
B.TECH IT
IIIT ALLAHABAD
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