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A $3-\text{ary}$ tree is a tree in which every internal node has exactly three children. Use induction to prove that the number of leaves in a $3-\text{ary}$ tree with $n$ internal nodes is $2(n-1)$.
asked in DS by Veteran (56.2k points)   | 156 views

1 Answer

+4 votes
No of nodes at level $i =3^i$

Let height of tree be $h$

So total no of internal nodes $=3^0+3^1+3^2+\dots +3^{h-1}=\frac{3^{h}-1}{2}$


No of leaf nodes $=3^h =2n+1=2(n-1)+3$

Let us prove by induction

Base case

$n=1$ (one internal node i.e., root node)

No of leaves $=2(1-1)+3=3$

Let it be true for $n$ internal nodes

Now we prove for $m$ nodes where $m=n+1$

We have $L(m)=2(n+1-1)+3$

Also $L(m)=L(n)+3-1\\=2(n-1)+3+3-1\\=2n+3$

So if $L(n)$ is true then $L(n+1)$ is also true

Hence proved by induction.
answered by Veteran (30.2k points)  
edited by

Apart from proving it by induction 

I was wondering does this formula hold correct ?


n = 2 L= 5

but by L = 2(n-1) = 1 ???????? How is it correct

let there be n internal nodes (including root)  and L leaves

then using degree 

(n-1)4 + 3 + L = 2( n+L-1)

L= 2n + 1

eg in above pic n = 2 so L = 5

(Just trying to learn calculating no of leaves - not worrying about proving by induction .....just want to know if formula is correct ? )

As per the wordings given in question your tree is correct. But i guess they meant a full ternary tree- otherwise no. of leaves cannot be a fixed value.
Can u plz draw a tree @arjun sir where the question's no. of leaf requirement will satisfy?

I m not getting any..
@Arjun ...i think question is wrong
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