GATE CSE 2002 | Question: 2.9

Question

The number of leaf nodes in a rooted tree of n nodes, with each node having $0$ or $3$ children is:

• A. $\frac{n}{2}$
• B. $\frac{(n-1)}{3}$
• C. $\frac{(n-1)}{2}$
• D. $\frac{(2n+1)}{3}$

Comments

Ternary tree given

Sum of degree $=2e$

$\underbrace{l}+\underbrace{(n-(l+1))}\times 4+\underbrace{1}\times 3=2e.....(1)$

# of leaf    # of internal        root node

 nodes        nodes

# of edges in a tree $'e'=n-1.......(2)$

Equating $'e'$ from both $(1)$ and $(2)$

$\dfrac{4n-3l-1}{2}=n-1$

$4n-3l-1=2n-2$

$\dfrac{2n+1}{3}=l$

$Ans: D$

---

Sum of degree =2e

Refer handshaking lemma

---

why are we multiplying #internal*4? and root node by 3?

---

Internal node has 3 childrens and 1 parent. That's why we multiplied by 4.

Root has 3 childrens and but no parents. Hence multiplied by 3.

Answer 1

$L =$ leaf nodes

$I =$ internal nodes

$n =$ total nodes $= L + I$

In a tree no. of edges  $= n - 1$

All edges are produced by only internal nodes so, 

$k\times I = n-1\qquad \to(1)$   (for $k-ary$ tree, in this question $k = 3$)

$L + I = n\qquad \to (2)$

Here, given options are in terms of "n". So, eliminating $I$ from $(1)$ and $(2)$,

$L = ((k-1)n+1)/k$

you get $L = (2n+1)/3$

Answer is D.

Comments

please explain why k*I = T-1 ??

---

since the tree is k-ary tree, each internal node will have k branches contributing total of k*I internal branches which is equal to total number of branches or edges in the tree which is n-1

Answer 2

A Different way to solve the same Question

Total no of nodes in rooted tree is n .And every node is going to have either 0 children or 3 children a/c to the question .

        

Total no of nodes	Leaf Nodes	n/2	(n-1)/3	(n-1)/2	(2n+1)/3
n = 4 (figure 1)	3	2	1	3/2	3
n = 7 (figure 2)	5	7/2	2	3	5
n = 10 (figure 3)	7	5	3	9/2	7
n = 13	9	13/2	4	6	9
n = 16	11	8	5	15/2	11

Option D satisfy all the condition of the question .So option D is the answer .

Comments

Yap, trial and error method.

Answer 3

total number of nodes (n)  = internal nodes(i )  +  leaf nodes(L) 

total number of nodes (n) =  3 * nodes with three children (x)  + 1 

n = 3*x + 1 

x = (n-1)/3

n = i +L

L = n-i

L = n - (n-1)/3

L = (2n+1)/3

Comments

total number of nodes (n) =  3 * nodes with three children (x)  + 1 

How'd you get this?

---

 Mizuki 

it is because             no of node with 3 children means internal node and every internal  node have 3 children  so total children is 3*nodes with three children 

so total node is= 3 * nodes with three children (x)  + 1 

why we add 1  ?

because the root node in not the child of any node which in not counted in 3*nodes with three children (x) 

---

Thanks! @Gurdeep Saini

Answer 4

We know no of leaf nodes(l) in a k-ary tree (l) = i(k-1)+1 (i represents internal nodes)

here k=3 so l=i(3-1)+1==>l=2i+1==>l-2i=1----------(a).

and we know total no of nodes in a tree (n) = Leaf nodes(l)+internal nodes(i)

so n=l+i;==>(b).

Add (a) & (b)

               (a)---->l-2i=1

          2*(b)---->2l+2i=2n

_________________________

3l=2n+1===> so l=(2n+1)/3;