Mumbai University > Computer Engineering > Sem 3 > Discrete Structures

Marks: 5 Marks

Year: May 2014

Let $\text{n=total number of leaves} \\ \text{i=internal nodes=n-20}$

$Then \hspace{2cm} \dfrac{n-1}{i}=m \ \ or \ \ n=mi+1 \\ But \hspace{3cm} i=\dfrac{n-1}{2} \Rightarrow n=2i+1 \\ \therefore \hspace{3.3cm} n=2(n-20)+1 \Rightarrow n = 39$

Hence, full binary three with 20 leaves has total of 30 nodes.