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Use structural induction to show that I(T) , the number of leaves of a full binary tree T , is 1 more than i (T) , the number of internal vertices of T .

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Prove that the number of leaves of a full binary tree T is 1 more than the number of internal vertices of T.

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01

Step 1:Introduction

Structural induction is used to prove that some proposition P(x) holds for all x of some sort of recursively defined structure, such as formulas, lists, or trees. A well-founded partial order is defined on the structures ("subformula" for formulas, "sublist" for lists, and "subtree" for trees).

02

Step 2:Solution

The objective is to use structural induction to show that t(T) , the number of leaves of a full binary trees T is more than i(T) , the number of internal vertices ofT.

Let the number of leaves of a full binary tree T be t(T) .

Let the number of internal vertices of T be i (T) .

We shall show that t(T) = 1 + i (T) , using structural induction.

03

Induction Step

Basis step:

Consider the tree consisting of a single vertex, say p.

Thus, there is no internal vertices and has leaf.

Therefore,

t(p) = 1 + i (p)

Recursive Step:

Let T1as its left sub tree and T2as its right sub tree and T2 .

We show that if the result is true for the trees T1and T2, then it is true for tree .

Now,

tT1=iT1+1tT2=iT2+1

The set of leaves of the tree T=T1T2is the union of the sets of eaves of T1and T2.

Thus, for the full binary tree T=T1T2,

ıT1T2=iT1+iT2+1

Add, the number of leaves,

tT1T2=tT1+tT2t(T)=iT1+1+iT2+1=tT1T2+1=i(T)+1tT2=iT2+1T=T1T2

Hence prove that the number of leaves of a full binary tree T is 1 more than the number of internal vertices of T .

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