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In a complete $k$-ary tree, every internal node has exactly $k$ children. The number of leaves in such a tree with $n$ internal node is:

1. $nk$
2. $(n-1)k + 1$
3. $n(k-1) +1$
4. $n(k-1)$
in DS
edited | 3.6k views
0

Before looking at answer just  form following two equation and try to solve -->

1. Equation for total degree.
2. Equation for total node.
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0
sum of degree$:l+(n-1)(k+1)+k=2e$

sum of degree$:l+kn+n-1=2e$

# of edges e$=\dfrac{l+kn+n-1}{2}$

and also # of edges in a tree $=(l+n)-1$

$\therefore \dfrac{l+kn+n-1}{2}=l+n-1$

$l=n(k-1)+1$

Answer :-> C) $n(k-1) +1$

Originally when we have root , there is only $1$ node, which is leaf.(There is no internal node.) From that "$+1$" part of formula comes from this base case.

When we $k$ children to nodes, we make root internal. So then Total Leaves $= n(k-1) + 1 = (k-1) + 1 = k$

In $k$ complete $k \ ary$ tree every time you add $k$ children , you add $k-1$ leaves.( $+k$ for leaves, $-1$ for node which you are attaching )

by Boss (41.9k points)
edited by
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Thanks for your explanation sir :-)
total nodes=nk+1(1 for root)

leaves =total nodes -internal nodes

=nk+1-n

=n(k-1)+1
by Boss (31.4k points)
+1
here k is not internal node ..... n is internal node and

leaves node = internal node (k-1)+1

leaves= n(k-1)+1

and total nodes = n(k-1)+1+n= nk+1

so ans should be (c)

http://www.geeksforgeeks.org/g-fact-42/

https://gateoverflow.in//1683/gate1998_2-11#viewbutton
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It was typo thanks for correcting... 0

@Pooja Palod

I've one doubt... How do u get total number of nodes=nk+1?

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@Pooja I couldn't get why you added 1 ? Is n't root is also an internal node ? So for n internal nodes having k children for each then it should be nk nodes in total ,why you adding 1 for root extra ,can you please explain ,thank you !

Option c.
by Active (3.3k points)

 leaves node = internal node (k-1)+1 leaves= n(k-1)+1 and total nodes = n(k-1)+1+n= nk+1 so ans should be (c) http://www.geeksforgeeks.org/g-fact-42/ https://gateoverflow.in//1683/gate1998_2-11#viewbutton
by Boss (17.1k points)
#leaves * degree_of_each_leave + #internal_nodes(excluding the root) *degree_of_each_internal node + degree of root=2(number of edges)

number of leaf node is( say l)

here degree of leaf node is 1

number o internal nodes(excluding root)is n-1

degree of each internal node is k+1

degree of root is k

number of edges=l+n-1

l+(n-1)(k+1)+k=2(l+n-1)

=>l+nk-k+n-1+k=2l+2n-2

=>l=n(k-1)+1
by Boss (14.4k points)
0
@Bhagi

more precisely just solve this two equation to get answer of any such ques!

N=ki+1

N=i+l

where  N=number of nodes i=no of internal node k= k-ary l=leaves
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in some que root node is not consider as internal node but in thi que root node is consider as internal node..why ??
+1 vote

To answer this type of question while practicing, always prefer derivation method instead of hit and trial

let total number of nodes =N

so N= total number of leaves nodes(say L) +total number  no internal nodes (n )

edge (E)= N-1  => E= n + L -1                     ............(1)

again edge (E)= (n* k) + (L*0)               ............(2) //{because only internal nodes are involved in forming an edge}

so from equation 1 and 2

n + L -1 = n*k

L=nk - n + 1

L=n(k-1)+ 1

by (457 points)