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+15 votes
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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. $n(k-1)$
asked in DS by Veteran (59.6k points)
edited by | 2.6k 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.
0

6 Answers

+12 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 )

answered by Boss (43k points)
edited by
0
Thanks for your explanation sir :-)
+12 votes
total nodes=nk+1(1 for root)

leaves =total nodes -internal nodes

            =nk+1-n

            =n(k-1)+1
answered by Boss (31.7k 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
0

It was typo thanks for correcting...

smiley

+2 votes
Option c.
answered by Active (3.3k points)
+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
answered by Boss (15.6k points)
+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
answered by Boss (14.3k 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
0
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

answered by (179 points)


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