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 sweetdivya@gmail.com wrote:
thanks a lot to all for their replies..
On 9 May 2010 11:23, rahul rai raikra...@gmail.com wrote:
can anyone give me links to
thanks a lot to all for their replies..
On 9 May 2010 11:23, rahul rai raikra...@gmail.com 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
aruntendulkar2...@gmail.com wrote:
This does not create a balanced tree
can anyone give me links to more educative and active groups like algogeeks
On Sun, May 9, 2010 at 2:11 AM, Arun prasath
aruntendulkar2...@gmail.com 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)
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 stdlib.h
#includeunistd.h
#include stdio.h
#define TEST2
#ifdef TEST1
int arr[] = { 1,2,3,4,5,6,7};
int max_elems
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,
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
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
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;
u are given a sorted lnked list construct a balanced binary search
tree from it.
--
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