Prove that the number of nodes with degree 2 in any Binary tree is 1 l

1.	Prove that the number of nodes with degree 2 in any Binary tree is 1 less than the number of leaves.
Answer» The proof is by induction on the size n of T . Let L(T) = No of leaves & D₂(T)= No of nodes of T of degree 2 To Prove D₂(T)=L(T)-1 Basic Case : n=1 , then T consists of a single node which is a leaf . L(T)=1 and D₂(T)=0 So D₂(T)=L(T)-1 Induction Step : Let n > 1 and assume for all non empty trees T1 of size k<n that D₂(T`)=L(T`)-1 Since n >1 , at least one of x=left(T) or y=right(T) is non empty Assume x is non empty ; the other case is symetric By the induction hypothesis , D₂(x) = L(y)-1 IF y is empty , then again by the induction hypothesis , D₂(y) =L(y)-1 and D₂(T) =D₂(x)+D₂(y)+1 =L(x)-1+L(y)-1+1 = L(x)+L(y)-1 D₂(T) = L(T)-1 Hence Proved

Prove that the number of nodes with degree 2 in any Binary tree is 1 less than the number of leaves.

Answer»

The proof is by induction on the size n of T .

Let L(T) = No of leaves & D₂(T)= No of nodes of T of degree 2

To Prove D₂(T)=L(T)-1

Basic Case : n=1 , then T consists of a single node which is a leaf .

L(T)=1 and D₂(T)=0

So D₂(T)=L(T)-1

Induction Step : Let n > 1 and assume for all non empty trees T1 of size k<n that D₂(T`)=L(T`)-1

Since n >1 , at least one of x=left(T) or y=right(T) is non empty

Assume x is non empty ; the other case is symetric By the induction hypothesis ,

D₂(x) = L(y)-1

IF y is empty , then again by the induction hypothesis ,

D₂(y) =L(y)-1 and

D₂(T) =D₂(x)+D₂(y)+1

=L(x)-1+L(y)-1+1

= L(x)+L(y)-1

D₂(T) = L(T)-1 Hence Proved

Discussion

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