Hi Atul,
You're correct. I was commenting on the original poster's words "I
guess its NlogN for balanced tree as it is traversing N nodes for H
times," which is not correct.
The general result is that the run time will be O(N) for any balance
rule where the ratio R of left to right subtree size satisfies 0 < Min
< R < Max < 1 for some Min and Max and for subtrees above a given
constant size.
If there is no such balance rule, then run time can be O(N^2).
This phenomenon is very handy to remember: The sum of any amount of
work that grows geometrically will be O(W) where W is the largest unit
of work done.
So for example when you are building an arbitrary-size stack class in
an array that resizes itself as needed, you should always grow and
shrink ( realloc() ) the array storage by a _factor_ greater than 1.
This way the copying costs remain O(N) where N is the number of stack
operations.
Gene
On Nov 21, 11:13 pm, atul anand <atul.87fri...@gmail.com> wrote:
> @Gene : i guess you are considering balanced tree.
>
> what if the tree is left skewed or right skewed .
>
> then height of thee will ne H=No. of node.
>
> so we will get :-
> 1+2+3+4....H
>
> so recurrence will be T(n) = T(n-1) + O(n)
>
> hence complexity will be O(N^2)
>
> correct me if i am missing somthing.
>
> Thanks in advance.
>
>
>
> On Sun, Nov 20, 2011 at 3:40 PM, Gene <gene.ress...@gmail.com> wrote:
> > My advice is don't guess. Do the math.
>
> > To print level L of a perfectly balanced tree, you will traverse 2^L
> > -1 nodes.
>
> > The outer loop that prints levels 1 .. H will therefore traverse
>
> > sum_{L=1..H} (2^L - 1) = 1 + 3 + 7 + ... 2^H - 1
>
> > Let N = 2^H - 1 be the number of nodes in the tree. Then
>
> > sum_{L=1..;H} (2^L - 1)
> > = sum_{L=1..H} (2^L) - H
> > = 2^(H+1) - 2 - H
> > = 2 N - log(N+1)
> > = O(N)
>
> > On Nov 20, 7:13 am, Ankuj Gupta <ankuj2...@gmail.com> wrote:
> > > I guess its NlogN for balanced tree as it is traversing N nodes for H
> > > times ie the height of tree which is logN for balanced and N for
> > > skewed tree. So it should be Nlogn and N^2
>
> > > On Nov 20, 9:27 am, tech coder <techcoderonw...@gmail.com> wrote:
>
> > > > @ sravanreddy001
> > > > complexity is O(N^2) whether tree is balanced or not doesn't matter
> > > > For each level it's visiting elements. all elements upto n-1 level .
> > > > i dont know from where u got the concept of logn , the code is not
> > > > making any decision to go in left or right , it is going in left and
> > right
> > > > both , so how it is nlogn.
>
> > > > On Sun, Nov 20, 2011 at 3:12 AM, sravanreddy001 <
> > sravanreddy...@gmail.com>wrote:
>
> > > > > Its NlogN if balanced.. Else N^2
>
> > > > > For each element it's visiting at most log N elements.(assuming
> > balanced)
>
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>
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