Cant we do something like this:
Make the middle element as root, middle element of the left side as its left
child and mid of the right half as its right child.....
and so on for the left and right subtrees.....................
it would bring out a balanced tree without rotations......
On Mon, May 3, 2010 at 5:41 PM, jalaj jaiswal <jalaj.jaiswa...@gmail.com>wrote:
> for simplicity in writin algo i've taken sorted array instead of list
> struct node * create( int *sorted,number of elements){
> struct node *temp,*left,*right;
> int tempii[100],tempiii[100];
> if(number of elemnts ==0)
> return NULL;
> temp->data=sorted[mid];
> temp->left=NULL;
> temp->right=NULL;
> if number of elements == 1
> return temp;
> int count=0;
> for(int i=0;i<mid;i++){
> tempii[i]=sorted[i];
> count++;
> }
> left=create(int *tempii,count);
> temp->left=left;
> count=0;
> for(i=mid+1;i<numberofelemnts;i++){
> tempiii[i]=sorted[i];
> count++;
> }
> right=create(int *tempiii,count);
> temp->right=right;
>
> return temp;
>
> }
>
> On Mon, May 3, 2010 at 5:36 PM, Rohit Saraf
> <rohit.kumar.sa...@gmail.com>wrote:
>
>> 1) Make the middle element the root.
>> Recursively make the left and right subtrees from the left and right
>> halves of the link list.
>>
>> 2) Implement balanced insertion in trees (via rotations on every step...).
>> Now insert each element
>> --------------------------------------------------
>> Rohit Saraf
>> Second Year Undergraduate,
>> Dept. of Computer Science and Engineering
>> IIT Bombay
>> http://www.cse.iitb.ac.in/~rohitfeb14
>>
>>
>>
>> On Sun, May 2, 2010 at 6:38 PM, divya <sweetdivya....@gmail.com> wrote:
>>
>>> u are given a sorted lnked list construct a balanced binary search
>>> tree from it.
>>>
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>
>
>
> --
> With Regards,
> Jalaj Jaiswal
> +919026283397
> B.TECH IT
> IIIT ALLAHABAD
>
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--
Regards,
Varun Bhatia
Research Intern
Advanced Development & Prototyping Group
Microsoft Research Lab India Pvt Ltd
Bangalore
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