Here is my version

Algorithm count(x)

1: if (x==nil || (left[x]== nil and right[x]==nil))
2:      return 0
3: return count(left[x]) + count(right[x]) +1

Time Complexity: O(n) where is n is total number of node in tree.

Thanks

On Wed, May 9, 2012 at 11:17 AM, Akshay Rastogi <akr...@gmail.com> wrote:

> you are not checking whether the current node is an internal node or not !!
>
> On Thu, May 3, 2012 at 12:47 AM, Rose <itro...@gmail.com> wrote:
>
>> Is this algorithm right or how shall I write it?
>>
>> *
>> *
>>
>> *
>> *
>>
>> *Construct an algorithm **Intern(**x**)**, which returns the number of
>> internal nodes in the tree. *
>>
>> * *
>>
>> Algorithm Intern(x)
>>
>>
>>
>> 1: if (x = nil) then
>>
>> 2:      return 0
>>
>> 3: else
>>
>> 4:      return 1 + Intern(left[x]) + Intern(right[x])
>>
>> 5: end if
>>
>>
>>
>>  *a   -    **Give the time complexity/Running time of your solution in **
>> O**-notation. *
>>
>> --
>> Med Venlig Hilsen/Kind regards
>>
>> Rose
>>
>>
>>
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>
>
>
> --
> AKSHAY RASTOGI
> BE(Hons) CS
> BITS PILANI , Pilani
>
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