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

  1. $\frac{n}{2}$
  2. $\frac{(n-1)}{3}$
  3. $\frac{(n-1)}{2}$
  4. $\frac{(2n+1)}{3}$
asked in DS by Veteran (68.8k points)
edited ago by | 1.9k views

4 Answers

+28 votes
Best answer

$L =$ leaf nodes

$I =$ internal nodes

$T =$ total nodes $= L + I$

in a tree no. of edges  $= T - 1$

all edges are produced by only internal nodes so 

$k*I = T-1$ ....................(1)   (for $k-ary$ tree, in this question $k = 3$)

$L + I = T$.....................(2)

solving $1$ and $2$ we get

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

So put $k = 3, T = n$

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

Ans D

answered by Veteran (13.8k points)
edited ago by
please explain why k*I = T-1 ??
+9 votes

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 .

answered by Veteran (44.3k points)
Yap, trial and error method.
+6 votes

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

 

 

answered by Active (2.4k points)
+2 votes

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;

 

 

answered by (367 points)


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