CS 103 Discrete Structures

Lecture 23

Trees (1)

Second Midterm Exam• 1st Lecture in December (same time as the

lecture)

• 75 minute duration

• Will cover all lectures delivered before the exam date

• Will consist of MCQ’s, fill-in-the-blanks, questions with short answers, writing of proofs, and drawing of diagrams

• If you miss this exam for any reason, you will have to appear for a makeup exam on the Thursday of the last week of teaching. That exam will cover all lectures delivered in the semester. It will consist of writing of proofs, drawing of diagrams and answering questions having 0.5-1 page answers.

TreesChapter 11

Chapter SummaryIntroduction to TreesApplications of TreesTree Traversal

Introduction to TreesSection 11.1

Section SummaryIntroduction to TreesRooted TreesTrees as ModelsProperties of Trees

Trees

A tree is an undirected, connected graph with no simple circuits

They are called trees because they look like trees. For example, family trees are graphs that represent genealogical chart

Trees as Models: ExamplesChemistry: Trees were introduced for counting the number of isomers of saturated hydrocarbons. The two isomers of butane are shown at the right

Computer Science: The organization of a computer file system into directories,subdirectories, and files is naturally represented as a tree

Business: Trees are usedto represent the structureof organizations

Which Graphs are Trees?Theorem: An undirected graph is a tree iff

there is a unique, simple path between any two of its vertices

G1 is a tree as it is a connected graphs with no

simple circuits

G2 is a tree as it is a connected graphs with no

simple circuits

G3 is not a tree as e, b, a, d, e is a simple circuit in

this graph

G4 is not a tree as it is not connected

Which Graphs are Trees?A connected graph that contains no simple circuits is a tree

Because a tree cannot have a simple circuit, a tree cannot contain multiple edges or loops. Therefore any tree must be a simple graph

What if there are no simple circuits but the graph is not connected? Each of theconnected componentsis a tree and the collectionis called a forest

Rooted TreesA rooted tree is a tree in which one vertex has been designated as the root and every edge is directed away from the root

We usually draw a rooted tree with its root at the top of the graph

We can change an unrooted tree into a rooted tree by choosing any vertex as the root

Different choices of the root produce different rooted trees

Different Roots, Different Trees

T with root a

T with root ca c

g e

d

gf

Tree T gf

d

c

c

f

Rooted TreesRoot

a b c

d e f a

b

cd e f

a

b

c

d

e

f

Rooted TreesWhat if a different root is chosen?

Roota b c

d e f

a

b c

d e f

a

b c

d

e f

a

b

c

d

e

f

Parent, Child & Sibling VerticesIf v is a vertex of rooted tree T other than the root, the parent of v is the unique vertex u such that there is a directed edge from u to v• The parent of a vertex is the vertex connected to

it on the path to the root• Every vertex except the root has a unique parent

When u is the parent of v, v is called the child of u• A child of a vertex v is a vertex of which v is the

parent

Vertices sharing a parent are called siblings

Parent, Child & Sibling Vertices

a

b c d

e f g h

i j k l m

Siblings

a is the parent of b, c, and d

e is the child of b

Ancestor & Descendent Vertices

The ancestors of a vertex (other than the root) are the vertices in the path from the root to this vertex, excluding the vertex itself and including the root

The descendants of a vertex v are those vertices that have v as an ancestor

Ancestors: Example

a

b c d

e f g h

i j k l m

k has 3 ancestors: f, d, a

Descendants: Example

a

b c d

e f g h

i j k l m

d has 6 descendants:f, g, h, k, l, m

LeavesA vertex with no children is called a leaf

a

b c d

e f g h

i j k l m

c, g, i, j, k, l and m are leaves

Internal VerticesVertices with children are called internal

vertices a

b c d

e f g h

i j k l m

a, b, d, e, f and h are internal vertices

Subtrees

If a is a vertex in a tree, the subtree with a as its root is the subgraph of the tree consisting of:

• a and its descendants, and

• all edges incident to these descendants

Subtrees: Example

a

b c d

e f g h

i j k l m

Subtree at b Subtree at d

m-ary and Full m-ary Trees

A rooted tree is called an m-ary tree if every internal vertex has no more than m children

A tree is called a full m-ary tree if every internal vertex has exactly m children

An m-ary tree with m 2 is called a binary tree

Full Binary Trees: ExampleWhat is the arity of this tree?

This is a binary (i.e.2-ary) tree

Is this a full m-ary tree?

Yes, this is a full binary tree since every internal vertex has exactly 2 children

Full 3-ary Trees: ExampleWhat is the arity of this tree?

This is a 3-ary tree

Is this a full m-ary tree?

Yes, this is a full 3-ary tree since every internal vertex has exactly 3 children

Full 5-ary Trees: ExampleWhat is the arity of this tree?

This is a 5-ary tree

Is this a full m-ary tree?

Yes, this is a full 5-ary tree since every internal vertex has exactly 5 children

3-ary Trees: ExampleWhat is the arity of this tree?

Some internal nodes have 2 children, but some have 3, so this is a 3-ary tree

Is this a full m-ary tree?

No. It is not a full-3-arytree, since oneinternal node hasonly 2 children

Ordered Rooted Tree

An ordered rooted tree is onewhere the children of eachinternal vertex are ordered

Ordered Rooted Tree: Left ChildIn an ordered binary tree, if an internal vertex has two children, then the left one is called the left child

f is the leftchild of d

a

b c

d e h i

f g j k l

m

a

b c

d e h i

f g j k l

mg is the rightchild of d

Ordered Rooted Tree: Right ChildIn an ordered binary tree, if an internal vertex has two children, then the right one is called the right child

Ordered Rooted Tree: Left Subtree

a

b c

d e h i

f g j k l

mLeft subtree of c

The subtree rooted at the left child of a vertex is called the left subtree

Ordered Rooted Tree: Right Subtree

a

b c

d e h i

f g j k l

m Right subtree of c

The subtree rooted at the right child of a vertex is called the right subtree

Level of a Vertex

Levels

a

b c d

e f g h

i j k l m

0

1

2

3

The level of a vertex in a rooted tree is the length of the unique path from the root to this vertex. The level of the root vertex is defined to be 0

Height of a TreeThe height of a rooted tree is the maximumof the levels of the vertices

a

b c d

e f g h

i j k l m

0

1

2

3

Height = 3

Levels

Level of a Vertex & Height of a Tree: Example

a is at level 0b, j, k are at level 1c, e, f, l are at level 2d, g, i, m, n are at level 3h is at level 4

Because the largestlevel of any vertex is 4,this tree’s height is 4

h

a

Properties of Trees

1. A tree with n vertices has n - 1 edges

2. A full m-ary tree with i internal vertices contains m · i + 1 vertices

3. There are at most mh leaves in an m-ary tree of height h

Balanced Trees

A rooted m-ary tree of height h is balanced if all leaves are at levels h or h - 1

Unbalanced Tree: Example

Section 11.1: Exercises1.a)Which vertex is the root?b)Which vertices are internal?c) Which vertices are leaves?d)Which vertices are

children of j?e)Which vertex is

the parent of h?f) Which vertices

are siblings of o?g)Which vertices are ancestors of m?h)Which vertices are descendants of b?i) What is the arity of this tree? Is it a full m-

ary tree?j) What is the level of each vertex?k)What is the height of the tree?l) Draw the subtrees rooted at a and c.

Section 11.1: Exercises17.How many edges does a tree with 200 vertices

have?

18.How many edges does a full binary tree with 500 internal vertices have?

19. Draw a tree having 6 vertices.

20. Draw a graph which has 6 vertices but is not a tree.

21. Draw an unbalanced tree having 6 vertices.

22. Draw a forest having 6 vertices.

CS 103 Discrete Structures

Lecture 24

Trees (2)

Second Midterm Exam• 1st Lecture in December (same time as the

lecture)

• 75 minute duration

• Will cover all lectures delivered before the exam date

• Will consist of MCQ’s, fill-in-the-blanks, questions with short answers, writing of proofs, and drawing of diagrams

• If you miss this exam for any reason, you will have to appear for a makeup exam on the Thursday of the last week of teaching. That exam will cover all lectures delivered in the semester. It will consist of writing of proofs, drawing of diagrams and answering questions having 0.5-1 page answers.

Applications of TreesSection 11.2

Section Summary Binary Search Trees Decision Trees Prefix Codes (Huffman Coding) Game Trees

Binary Search TreesSearching for items in a list is an important task

The goal of search is to efficiently find the required item in an ordered collection

A binary search tree is a good tool for this purpose

A binary search tree is binary tree in which each child of a vertex is designated as the left or right child, and each vertex is labeled with a key, which is one of the search itemsVertices are assigned keys so that key of a vertex is both

larger than the keys of all vertices on its left subtree and smaller than the keys of all vertices in the right subtree

Binary Search Trees: ExampleFind a binary search tree for the words mathematics, physics, geography, zoology, meteorology, geology, psychology, and chemistry using alphabetical ordering

Decision TreesRooted trees can be used to model a

decision process in which a sequence of decision leads to a solution

A rooted tree in which each internal vertex corresponds to a decision, with a subtree at these vertices for each possible outcome of the decision, is called a decision treeThe possible solutions of the problem

correspond to the paths to the leaves of this rooted tree

Decision Trees: ExampleSuppose there are 7 identical coins, and a counterfeit coin, that weighs less than the others. How many weighings are necessary using a balance scale to determine which of the 8 coins is the counterfeit one?

Using Bit-Strings to Encode Text

Consider the problem of data compression. Suppose that we have a 100,000-character file that contains only 6 different characters, a, b, c, d, e, and f. Some of these characters occur more frequently than others

Currently, the file is stored using a fixed-length code of 3 bits per character, where:

a = 000 , b = 001, c = 010, d = 011, e = 100, f = 101

This requires 300,000 bits to store the file

Prefix CodesWe can use Huffman coding which uses a variable-length code to compress the file

We can use a 0 to represent the most frequently-occurring letter in the file, which will save us two bits per occurrence

Huffman codes are prefix codes, which means that all bit patterns are unambiguous, i.e., no bit-string corresponds to more than one sequence of letters. This means that bit-string of one letter never occurs as the first part of the bit-string for another letter

Using Huffman coding, we can store the file in 224,000 bits

Prefix Codes: Examples10

1

1

1

1

0

0

0

0

0 = a

10 = b

110 = c

1110 = d11110

= e11111

= f

10

1

10

1

10

0 = a

101 = b

100 = c

111 = d

1101 = e

1100 = f

0 0

Huffman Codes: Codewordsa b c d e f

Frequency of occurrence (in thousands)

45 13 12 16 9 5

Codeword (fixed-length)

000

001

010

011 110 101

Codeword(variable length)

0 101

100

111

1101

1100

Huffman Codes are Unambiguous

Since we are using a prefix code, we can simply concatenate our codewords together to produce our bit-string

Example: The string abc can be represented as 0101100

This is unambiguous. That is, this character-string and bit-string pairing is unique

Huffman Code DecodingOnly one character can begin with 0; that is a. So a must be the first character in our string

This leaves 101100. The next bit is a 1; five characters can begin with 1, so we look at the second bit. Two characters can begin with 10, so we look at the third bit. Only one character can begin with 101; that is b

This leaves 100. Again, looking at all three bits, we see that this character must be c

Game TreesTrees can be used to model and analyze certain2-player games like tic-tack-toe

In game trees, vertices represent the position that a game can be in as it progresses

The edges represent legal moves between these positions

The root represents the starting point

When the game is at position represented by a vertex at an even level, it is the first player’s move. Otherwise, it is the second player’s move

The leaves of the game tree represent the concluding positions of a game

Game Tree Example: Tic-Tac-Toe

Tree TraversalSection 11.3

Section SummaryTraversal AlgorithmsInfix, Prefix, and Postfix Notation

Tree Traversal

Procedures for systematically visiting every vertex of an ordered tree are called traversals

The three most commonly used traversals are:1. Preorder traversal2. Inorder traversal3. Postorder traversal

Preorder TraversalLet T be an ordered rooted tree with root r:1.If T consists only of r, then r is the preorder

traversal of T2.Otherwise, suppose that T1, T2, …, Tn are the

subtrees of r from left to right in T. The preorder traversal begins by visiting r, andcontinues by traversingT1 in preorder, then T2

in preorder, andso on, until Tn istraversed inpreorder

Preorder Traversal (contd.)

procedure preorder (T: ordered rooted tree)r := root of Tlist rfor each child c of r from left to right T(c) := subtree with c as root preorder(T(c))

Inorder TraversalLet T be an ordered rooted tree with root r:1.If T consists only of r, then r is the inorder

traversal of T2.Otherwise, suppose that T1, T2, …, Tn are the

subtrees of r from left to right in T. The inorder traversal begins by traversing T1

in inorder, then visiting r,and continues bytraversing T2

in inorder, andso on, until Tn

is traversedin inorder

Inorder Traversal (contd.)

procedure inorder (T: ordered rooted tree)r := root of Tif r is a leaf then list relse l := first child of r from left to right T(l) := subtree with l as its root inorder(T(l)) list(r) for each child c of r from left to right T(c) := subtree with c as root inorder(T(c))

Postorder TraversalLet T be an ordered rooted tree with root r:1. If T consists only of r, then r is the postorder

traversal of T2.Otherwise, suppose that T1, T2, …, Tn are the

subtrees of r from left to right in T. The postorder traversal begins by traversing T1

in postorder, then T2

in postorder, andso on, after Tn

is traversedin postorder,r is visited

Postorder Traversal (contd.)

Procedure postordered (T: ordered rooted tree)r := root of Tfor each child c of r from left to right T(c) := subtree with c as root postorder(T(c))list r

Expression TreesComplex expressions (e.g. compound propositions, set combinations, arithmetic expressions) can be represented using ordered rooted tree

Example:A binary tree representing((x + y) ↑ 2 ) + ((x − 4)/3)

Expression Forms: Infix, Prefix, PostfixInfix expression form has operators written between operands. Parentheses are sometimes added to infix expressions to remove ambiguity. Example: x + y

Prefix expression form (or Polish notation form) has operators written before the operands. Parentheses are never needed for expressions in the prefix form. Example: + x y

Postfix expression form (or reverse Polish notation form) has operators written after the operands. Parentheses are never needed for expressions in the postfix form. Example: x y +

Note: All three example expressions shown above lead to the exact same result, i.e. the sum of x and y

Infix NotationAn inorder traversal of the tree representing an expression produces the original expression when parentheses are included

Trees shown bellow all lead to the same infix expression x + y/x + 3

To make such expressions unambiguous, it is necessary to include parentheses in the inorder traversal whenever we encounter an operation. The fully parenthesized expression obtained this way is said to be in the infix form

(x + (y/x)) + 3 x + (y/(x + 3))(x + y)/(x + 3)

Infix NotationWe obtain the infix form of an expression by

traversing its binary tree in inorder

x + y/x + 3 x + y/x + 3x + y/x + 3 (x + (y/x)) + 3 x + (y/(x + 3))(x + y)/(x + 3)

Trees shown bellow all lead to the same infix expressionx + y/x + 3. To make infix expressions unambiguous, it is necessary to include parentheses in the inorder traversal whenever we encounter an operation

Prefix NotationWe obtain the prefix form of an

expression by traversing its binary tree in preorder

Prefix form of ((x + y) ↑ 2 ) + ((x − 4)/3) is + ↑ + x y 2 / − x 4 3

Prefix expressions are evaluated by working from right to left. When we encounter an operator, we perform the corresponding operation with the two operations to the right

Example: We show the steps used to evaluate a particular prefix expression

Postfix NotationWe obtain the postfix form of an

expression by traversing its binary tree in postorder

Postfix form of ((x + y) ↑ 2 ) + ((x − 4)/3) is x y + 2 ↑ x 4 − 3 / +

A binary operator follows its two operands. So, to evaluate an expression one works from left to right, carryingout an operation represented by an operator on its preceding operands

Example: We show the steps used to evaluate a particular postfix expression