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Lecture Notes on Lists II and Binary Trees (COMP11120) $7.58   Add to cart

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Lecture Notes on Lists II and Binary Trees (COMP11120)

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Enhance your understanding of advanced list operations and binary trees with these comprehensive lecture notes for COMP11120. Covering essential topics such as list manipulation, traversal techniques, binary tree structures, and tree traversal algorithms, these notes are designed to provide you wit...

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  • May 30, 2024
  • 2
  • 2023/2024
  • Class notes
  • Andrea schalk
  • More on lists, and introduction to binary trees
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More on Lists, and Introduction to Binary Trees

Binary Trees

Terminology
L root
in a
binary tree every mode >
- in a full binary tree
, every node


7
has at most 2 children
. ·
is either a leaf

leaves & · or has a left and right child , which themselves are the root of a tree.


full binary tree
T 7 in a
every node

has 0 or 2 children
. - a full binary tree consists of


· either just a root mode


· or a root and a left and a right (subtree.




Full Binary Trees
Definition A full binary tree with labels from set S base Step case
: a is given
by case




basecase trees where seS tree S 7 S treeg (t t) ,
> S
,



Step case Given trees t ,
t with labels from S
, t t

treeg (t ,
tl) is a tree with labels from S
.




Example : trees (treen (troes treez) , , treed


3
#
*
-
H 04
#

3 28





FBTreess : set of all full binary trees with labels from S
.



Example Define
: operation FBTreess <
IN that returns the height of Example : Give an operation no :
FiTreess > 11I that returns the
tree number of nodes tree
a .

of a .




base case Light (trees) = D base case noltroos) =
1

Step case Light (trees (t t)) ,
= 1 + max light , light ↑
Step case no(troeg(t ,
t) = not + not + 1




Example : Show that for all to FBTroesg ,
light not Example : Show that for te FBTreesg ,
not is old




base case Light trees = 0 1 = no trees base case noltrees) = 1 which is odd


Step case
hight (treeg(t t ,
=
1 + max (lightt lightt) , ind .
hyp not , not' are odd
max(a, b) ( a + b

& 1 +
light + lightt Step case nol trees (t t) = 1 + not + not
,


not not E *

>
- 1 + + indhyp by ind .
hyp . We can find i
,
je IN such that

= no (trees (t , tl) not = 2i + 1


not =
2j + 1

L


= 1 + 2i + 1 + 2j1
= 1 + 2(i + j +
1)

=
1 + even


= odd

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