Answer is 5.
the relation is
No of leaf nodes = (n-1)*(no of internal nodes) + 1

Paul

On Thu, Aug 11, 2011 at 10:44 PM, amit karmakar
<[email protected]>wrote:

> * correction
> (5*5(There are 5*5 nodes in level 2)-4(These became internal nodes..))
>
> On Aug 11, 9:58 pm, amit karmakar <[email protected]> wrote:
> > 5 is possible.
> > Considering root of the tree to be at level 0,
> > level 1 and level 2 are completely filled.
> >
> > There are 5 internal nodes in level 1, (since all level 2 nodes are
> > present)
> > Now only (10 - 5(from level 1)+1(the root)) nodes are required.
> > So choose 4 nodes from level 2 and make them interior node.
> > So you get 4*5(4 nodes have 5 children) + (5*5(There are 5*5 nodes in
> > level 2)-4(These became leaves)) leaves.
> >
> > Unfortunately 5 is not in the option
> >
> > On Aug 11, 7:31 pm, rShetty <[email protected]> wrote:
> >
> >
> >
> > >  A complete n- array tree in which each node has n children or no
> > > children, let i be the number of internal nodes and L be the number of
> > > leaves in a complete n- array tree. If L=41 and i=10 what is the value
> > > of n.
> >
> > > a. 3    b. 6   c. 4
> >
> > > How to solve such problems??
>
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