the best approach to it is to build a balanced tree from bottom to up rather
than top to bottom.
On 12 May 2010 22:47, divya jain wrote:
> thanks a lot to all for their replies..
>
>
> On 9 May 2010 11:23, rahul rai wrote:
>
>> can anyone give me links to more educative and active groups like
>
thanks a lot to all for their replies..
On 9 May 2010 11:23, rahul rai wrote:
> can anyone give me links to more educative and active groups like algogeeks
>
> On Sun, May 9, 2010 at 2:11 AM, Arun prasath
> wrote:
> > This does not create a balanced tree but ensures that every element in
> the
can anyone give me links to more educative and active groups like algogeeks
On Sun, May 9, 2010 at 2:11 AM, Arun prasath
wrote:
> This does not create a balanced tree but ensures that every element in the
> tree is accessible by lg(n) time.
>
> Time : Complexity O(n)
>
>
> [a...@91blore-srv1 ~]
This does not create a balanced tree but ensures that every element in the
tree is accessible by lg(n) time.
Time : Complexity O(n)
[a...@91blore-srv1 ~]$ cat recursion.c
#include
#include
#include
#define TEST2
#ifdef TEST1
int arr[] = { 1,2,3,4,5,6,7};
int max_elems = sizeof(arr)/sizeof(ar
i guess the rotation solution i gave will take O(n) with the list as
well
(btw.. u can convert a list to array :P)
--
Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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, Ma
it is easy to traverse the array. as u can directly reach middle element in
constant time. but for linked list u cant do dat. so the time complexity of
both the solution given above are O(nlogn). is there any other better
complexity solution.
On 3 May 2010 17:41, jalaj jaiswal wrote:
> for simpl
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[
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
u are given a sorted lnked list construct a balanced binary search
tree from it.
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to algoge...@googlegroups.com.
To unsubscribe from this group, send email to
a
10 matches
Mail list logo