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

- Equation for total degree.
- Equation for total node.

15 votes

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:

- $nk$
- $(n-1)k + 1$
- $n(k-1) +1$
- $n(k-1)$

15 votes

Best answer

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 )

19 votes

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

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

0

@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 !

2 votes

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 |

2 votes

#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

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

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**