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27 Example 2: Full Binary Trees
The set of full binary trees can be defined as follows:
Basis: There is a full binary tree consisting of a single vertex r
Induction: If T1 and T2 are disjoint full binary trees and r in A
is a node, then <T1, r, T2 > is a full binary tree with root r and
left subtree T1 and right subtree T2. The difference between binary trees and full binary trees is in
the basis step. A binary tree is full if and only if each node is
either a leaf or has precisely two children. 28 Small Full Binary Trees
Level 0: Level 1: not full!
Level 2: 29 Height of a Full Binary Tree Let T be a full binary tree over an alphabet A. We define the
height h(T) of a full binary tree as follows:
Basis: For r in A, we define h(r) = 0; that is, the height of a full
binary tree with just a single node is 0.
Induction: If L and R are full binary trees and r in A, then the
tree <L,r,R> has height
h( <L,r,R> ) = 1 + max( h(L), h(R) ). 30 Number of Nodes of a Full Binary Tree Let n(T) denote the number of nodes of a full
binary tree over an alphabet A. Then
Basis: For r in A, we have n(r)=1.
Induction: If L and R are full binary trees and r
in A, then the number of nodes of <L,r,R> is given
by
n(<L,r,R>) = 1+n(L)+n(R).
31 Example of Structural Induction
Theorem: Let T be a full binary tree over an alphabet A. Then
we have n(T) <= 2h(T)+11.
Proof: By structural induction.
Basis step: For r in A, we have n(r)=1 and h(r)=0, therefore, we
have n(r) = 1 <= 2(0+1)1 = 2h(r)+11, as claimed.
Inductive step: Suppose that L and R are full binary trees that
satisfy n(L) <= 2h(L)+11 and n(R) <= 2h(R)+11. Then the tree
T=<L,r,R> satisfies:
n(T) = 1 + n(L) + n(R) <= 1 + 2h(L)+11 + 2h(R)+11
<= 2 max(2h(L)+1, 2h(R)+1)1, since a+b <= 2max(a,b)
<= 2.2max(h(L),h(R))+1 1=2.2h(T)1 = 2h(T)+11
32...
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This note was uploaded on 03/24/2014 for the course CSCE 222 taught by Professor Math during the Fall '11 term at Texas A&M.
 Fall '11
 math
 Recursion

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