BCT 2083 DISCRETE STRUCTURE AND APPLICATIONSBCT 2083 DISCRETE STRUCTURE AND APPLICATIONS

CHAPTER 4CHAPTER 4TREESTREES

SITI ZANARIAH SATARISITI ZANARIAH SATARI

FIST/FSKKP UMP I0910FIST/FSKKP UMP I0910

CHAPTER 4 TREES

CONTENTCONTENT

4.1 Introduction to Trees4.2 Application of Trees4.3 Tree Traversal4.4 Trees and Sorting (Data

structure)

4.5 Spanning Trees 4.6 Minimum Spanning

Trees

4.1 INTRODUCTION TO TREES4.1 INTRODUCTION TO TREES

• Define and recognize a tree

• Define and recognize a rooted tree

• Define and recognize a m-ary tree

CHAPTER 4 TREES

4.1 INTRO

DUCTIO

N TO

TREES

►A tree is a connected undirected graph with no simple circuit.

►An undirected graph is a tree if and only if there is a unique simple path between any two of its vertices.

►THEOREM: A tree with n vertices has n – 1 edges.

A TreeA Tree

a b c

d e f

a b c

d e f

a b c

d e f

NOT A TREE NOT A TREE A TREE

EXAMPLES

4.1 INTRO

DUCTIO

N TO

TREES

1. Which of the following graphs are trees?

EXERCISE 4.1EXERCISE 4.1

a b c

d eFIGURE 1

a b

c d e

FIGURE 2

a

b

c d

e

FIGURE 3 FIGURE 4

v1

v2

v4

v3

v5

4.1 INTRO

DUCTIO

N TO

TREES

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

► Different choice of root produce different rooted tree

A Rooted TreeA Rooted Tree

a

b

c

d

e

f a

b c d

ef

a

b

c

d

e

f

A TREE A TREE with root c

A TREE with root a

EXAMPLES

g

gg

4.1 INTRO

DUCTIO

N TO

TREES

• Parent – A vertex other than root is a parent if it has one or more children The parent of c is b

• Children – If A is a vertex with successors B and C, then B and C are the children of A. The children of a is b, f and g

• Siblings – Children with the same parent vertex. h, i and j are siblings

• Level – the length of the unique path from the root to a vertex Vertex a is at level 0 Vertices d and e is at level 3

Properties of Rooted TreesProperties of Rooted Trees

a

b f g

h

d

EXAMPLES

e

ijc

k l m

• Height – The maximum level of all the vertices The height of this

tree is 3.

4.1 INTRO

DUCTIO

N TO

TREES

• Ancestor of a vertex (v) – the vertices in the path from the root to this vertex excluding this vertex. The ancestors of e are c, b and a

• Descendent of a vertex (v) – vertices that have v as ancestor. The descendants of b are c, d and e

• Leaf – A vertex with no children The leaves are d, e, f, i, k, l and m

• Internal Vertices – vertices that have children The internal vertices are a, b, c, g,

h and j

Properties of Rooted TreesProperties of Rooted Trees

a

b f g

h

d

EXAMPLES

e

ijc

k l m

A subtree rooted at g

• Subtree – A subgraph of the tree consisting of a root and its descendent and all edges incident to these descendent.

4.1 INTRO

DUCTIO

N TO

TREES

2. Answer these questions about the rooted tree illustrated.

EXERCISE 4.1EXERCISE 4.1

a) Which vertex is the root?b) Which vertices are internal?c) Which vertices are leaves?d) Which vertices are children of g?

e) Which vertex is the parent of o?f) Which vertices are siblings of e?g) Which vertices are ancestors of m?h) Which vertices are descendants of d?

a

bc

d

h

j k

r

if

no

p

e g

l m

q s

e) Find a subgraph.f) Find the level of

each vertex.g) What is the height

of this tree?h) Is this tree a

balanced tree?

4.1 INTRO

DUCTIO

N TO

TREES

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

► The tree is called a full m-ary tree if every internal vertex has exactly m children.

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

mm-ary Tree-ary Tree

a

b c d

f

a

b

c

d

e

f

A full binary TREEA 3-ary TREE

EXAMPLES

g

g

a

b c d

ef g

A full 4-ary TREE

e

h i j k l m

4.1 INTRO

DUCTIO

N TO

TREES

► THEOREM 1: A full m-ary tree with i internal vertices contains n = mi + 1 vertices

► THEOREM 2: A full m-ary tree with

i. n vertices has internal vertices and leaves.

ii. i internal vertices has vertices and leaves.

iii. l leaves has vertices and internal vertices.

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

mm-ary Tree-ary Tree

1nim

1 1m n

lm

1n mi 1 1l m i

11

mlnm

11

lim

hm

4.1 INTRO

DUCTIO

N TO

TREES

Suppose that someone starts a chain letter. Each person who receives the letter is asked to send it on to four other people. Some people do this, but others do not send any letters.

How many people have seen the letter, including the first person, if no one receives more than one letter and if the chain letter ends after there have been 100 people, who read it but did not send it out?

EXAMPLE: EXAMPLE: mm-ary tree-ary tree

Solution:

Send letter to four other people 4-ary tree ⇛ ⇛ m = 4People send out the letter internal vertices ⇛ ⇛ iPeople do not send out the letter leaves ⇛ ⇛ l = 100

By theorem 2 (iii):Number of people have seen this letter ⇛ n = (4·100 – 1) / (4 – 1) = 133Number of people send out the letter ⇛ i = 133 – 100 = 33

4.1 INTRO

DUCTIO

N TO

TREES

EXERCISE 4.1EXERCISE 4.1

3. A chain letter starts with a person sending a letter out to 10 others. Each person is asked to send the letter out to 10 others, and each letter contains a list of previous six people in the chain. Unless there are fewer than six names in the list, each person sends one dollar to the first person from the list, moves up each of other five names one position, and insert his names at the end of this list.

If no person breaks the chain and no one receives more than one letter, how much money will a person in the chain ultimately receive?

4.1 INTRO

DUCTIO

N TO

TREES

EXERCISE 4.1 : EXTRAEXERCISE 4.1 : EXTRA

PAGE : 693, 694 and 695

Rosen K.H., Discrete Mathematics & Its Applications, (Seventh Edition), McGraw-Hill, 2007.

4.24.2 APPLICATION OF TREESAPPLICATION OF TREES

• Introduce Binary Search Trees

• Introduce Decision Trees

• Introduce Game Trees

CHAPTER 4 TREES

4.2 APPLICATION

OF TREES

• A binary tree in which each child of a vertex is designated as a right or left child

• No vertex has more than one right child or left child

• Each vertex is labeled with a key

• Vertices are assigned keys so that the key of a vertex is both larger than the keys of all vertices in its left subtree and smaller than the keys of all vertices in its right subtree.

Binary Search TreesBinary Search TreesEXAMPLE

Binary search tree for the words mathematics, physics, geography, zoology,

meteorology, geology, psychology, and chemistry using alphabetical order

geography

chemistry

mathematics

geology

physics

metereologyzoology

psycology

4.2 APPLICATION

OF TREES

• A rooted tree in which each internal vertex corresponds to a decision, with a subtree at these vertices for each possible outcome of decision.

• The possible solutions of the problem correspond to the paths to the leaves of this rooted tree.

Decision TreesDecision TreesEXAMPLE

A Decision tree that orders the elements of the list a, b, c

a : b

a : c b : c

b : c c > a > b

a > b a < b

a > c a < c

a : c c > b > a

b > c b < c

a > c > b

b > c b < c

a > b > c b > c > a

a > c a < c

b > a > c

4.2 APPLICATION

OF TREES

• Use trees to analyze certain types of games • Vertices represent the positions can be in as it progresses• Edges represent legal moves between this position • Games tree is infinite if the games they represent never end

Game TreesGame Trees

EXAMPLE

A Game tree that represents

the first two level of the

tic-tac-toe game.

4.2 APPLICATION

OF TREES

EXERCISE 4.2 : EXTRAEXERCISE 4.2 : EXTRA

PAGE : 708, 709 and 710

Rosen K.H., Discrete Mathematics & Its Applications, (Seventh Edition), McGraw-Hill, 2007.

4.34.3 TREE TRAVERSALTREE TRAVERSAL

• Determine preorder traversal, inorder traversal and postorder traversal of an ordered rooted tree.

• Determine the infix, prefix, and postfix form of an expression.

CHAPTER 4 TREES

4.3 TREE TRAVERSAL

• Ordered trees are often used to restore data/info.

• Tree traversal is a procedure for systematically visiting each vertex of an ordered rooted tree to access data.

• If the tree is label by Universal Address System we can totally order the vertices using lexicographic ordering

– Example: 0 < 1 < 1.1 < 1.2 < 1.2.1 < 1.3 < 2 < 3 < 3.1 < 3.1.1 < 3.1.2 < 3.1.2.1 < 3.1.2.2 < 4 < 4.1

• Tree traversal algorithm– Preorder, inorder and postorder traversal

Tree TraversalTree Traversal

4.3 TREE TRAVERSAL

• Let T be an ordered rooted tree with root r.

– If T consists only of r, then r is the preorder traversal of T.

– If T1, T2, …, Tn are subtrees at r from left to right in T, then the preorder traversal begins by visiting r, continues by traversing T1 in preorder, then T2 in preorder, and so on until Tn is traversed in preorder.

Preorder TraversalPreorder Traversal

Tn

r STEP 1 Visit r

T2 …

STEP 2 Visit T1 in preorder

T1

STEP n+1 Visit Tn in preorder

STEP 3 Visit T2 in preorder

TIPS

Preorder Traversal: Visit root, visit

subtrees left to right

4.3 TREE TRAVERSAL

EXAMPLE: Preorder TraversalEXAMPLE: Preorder Traversalb

cd

hj k

i

n p

e g

l m

a

o

f

b d

hj k

i

n p

eg

l m

a•

o

f

c•

j k

n po

g

l m

ea•

c•

b•

d•

f•

h•

i•

a•

c•

b•

d•

f•

h•

i•

e•

j•

k

n po

l•

m•

a•

b•

e•

j•

k•

n•

o•

p•

c•

d•

h•

i•

l•

m•

f•

The preorder traversal of T

T

ITERATION 1

ITERATION 2

ITERATION 3

ITERATION 4

g•

g•

4.3 TREE TRAVERSAL

• Let T be an ordered rooted tree with root r.

– If T consists only of r, then r is the inorder traversal of T.

– If T1, T2, …, Tn are subtrees at r from left to right in T, then the inorder traversal begins by traversing T1 in inorder, then visiting r, continues by traversing T2 in inorder, and so on until Tn is traversed in inorder.

Inorder TraversalInorder Traversal

Tn

r STEP 2 Visit r

T2 …

STEP 1 Visit T1 in inorder

T1

STEP n+1 Visit Tn in inorder

STEP 3 Visit T2 in inorder

TIPS

Inorder Traversal: Visit leftmost subtree, Visit root,

Visit other subtrees left to

right.

4.3 TREE TRAVERSAL

EXAMPLE: Inorder TraversalEXAMPLE: Inorder Traversalb

cd

hj k

i

n p

e g

l m

a

o

f

b d

hj k

i

n p

eg

l m

a•

o

f

c•

j k

n po

g

l m

e f•

c•

b•

d•

a•

h•

i•

a•

l•

b•

g•

f•

h•

i•

e•

j•

k

n po

m•

d•

j•

e•

n•

k•

o•

p•

f•

a•

g•

m•

h•

i•

d•

b•

l•

The inorder traversal of T

T

ITERATION 1

ITERATION 2

ITERATION 3

ITERATION 4

c•

c•

4.3 TREE TRAVERSAL

• Let T be an ordered rooted tree with root r.

– If T consists only of r, then r is the postorder traversal of T.

– If T1, T2, …, Tn are subtrees at r from left to right in T, then the preorder traversal begins by traversing T1 in postorder, then T2 in postorder, and so on until Tn is traversed in postorder and ends by visiting r.

Postorder TraversalPostorder Traversal

Tn

r STEP n+1 Visit r

T2 …

STEP 1 Visit T1 in postorder

T1

STEP n Visit Tn in postorder

STEP 2 Visit T2 in postorder

TIPS

Postorder Traversal: Visit subtrees left to right, Visit

root.

4.3 TREE TRAVERSAL

EXAMPLE: Postorder TraversalEXAMPLE: Postorder Traversalb

cd

hj k

i

n p

e g

l m

a

o

f

b d

hj k

i

n p

eg

l m

a•

o

f

c•

j k

n po

g

l m

e a•

c•

b•

d•

f•

h•

i•

a•

c•

b•

d•

f•

h•

i•

e•

j•

k

n po

l•

m•

j•

n•

o•

p•

k•

e•

f•

b•

l•

m•

d•

a•

h•

i•

c•

The preorder traversal of T

T

ITERATION 1

ITERATION 2

ITERATION 3

ITERATION 4

g•

g•

4.3 TREE TRAVERSAL

1. Determine the order in which a preorder, inorder, and postorder traversal visits the vertices of the following rooted tree.

EXERCISE 4.3EXERCISE 4.3

a

b f g

h

d e

ijc

k l m

a

bc

d

h

j k

r

if

no

p

e g

l m

q sFIGURE 1 FIGURE 2

4.3 TREE TRAVERSAL

• We can represent complicated expression (propositions, sets, arithmetic) using ordered rooted trees.

• EXAMPLE: A binary tree representing ((x+y)↑2)+((x-4)/3)

Represent Expression by Rooted TreeRepresent Expression by Rooted Tree

↑

+

+

2

/

_ 3

x y x 4

4.3 TREE TRAVERSAL

• We obtain the Infix form of an expression when we traverse its rooted tree in Inorder.– The infix form for expression ((x+y)↑2)+((x-4)/3) is x + y ↑2 + x – 4 / 3 or ((x+y)↑2)+((x-4)/3)

• We obtain the Prefix form of an expression when we traverse its rooted tree in Preorder.– The prefix form for expression ((x+y)↑2)+((x-4)/3) is

+ ↑ + x y 2 / - x 4 3

• We obtain the Postfix form of an expression when we traverse its rooted tree in Postorder.– The postfix form for expression ((x+y)↑2)+((x-4)/3) is

x y + 2 ↑ x 4 – 3 / +

Infix, Prefix & Postfix NotationInfix, Prefix & Postfix Notation

4.3 TREE TRAVERSAL

• Working right to left and performing operations using the operands on the right.

• EXAMPLE: The value of the prefix expression + - * 2 3 5 / ↑ 2 3 4 is 3

Evaluating Prefix ExpressionEvaluating Prefix Expression

+ - * 2 3 5 / ↑ 2 3 42 ↑ 3 = 8

+ - * 2 3 5 / 8 4 8/4 =2

+ - * 2 3 5 22 * 3 = 6

+ - * 2 3 5 22 * 3 = 6

+ - 6 5 26 – 5 =1

+ 1 21 + 2 = 3

4.3 TREE TRAVERSAL

• Working left to right and performing operations using the operands on the left.

• EXAMPLE: The value of the postfix expression 7 2 3 * - 4 ↑ 9 3 / + is 4

Evaluating Postfix ExpressionEvaluating Postfix Expression

7 2 3 * - 4 ↑ 9 3 / +2 * 3 = 6

7 - 6 = 1

1 ↑ 4 = 1

9/3 = 3

1 + 3 = 4

7 6 - 4 ↑ 9 3 / +

1 4 ↑ 9 3 / +

1 ↑ 4 = 1

1 4 ↑ 9 3 / +

1 9 3 / +

1 3 +

4.3 TREE TRAVERSAL

EXERCISE 4.3EXERCISE 4.3

2. Represent the following expression using binary trees. Then write these expression in infix, prefix and postfix notations.

a. ((x+2)↑3)*(y – (3+x)) – 5b. (A∩B) – (A∪(B – A))

3. What is the value of these expression in prefix expression?a. + – ↑3 2 ↑ 2 3 / 6 – 4 2b. * + 3 + 3 ↑ 3 + 3 3 3

4. What is the value of these expression in postfix expression?a. 3 2 * 2 ↑ 5 3 – 8 4 / * –b. 9 3 / 5 + 7 2 – *

4.3 TREE TRAVERSAL

EXERCISE 4.3 : EXTRAEXERCISE 4.3 : EXTRA

PAGE : 722, 723 and 724

Rosen K.H., Discrete Mathematics & Its Applications, (Seventh Edition), McGraw-Hill, 2007.

4.54.5 SPANNING TREESSPANNING TREES

• Find spanning trees of a simple graph

CHAPTER 4 TREES

4.5 SPANN

ING TREES

• Let G be a simple graph. A spanning tree of G is a subgraph of G that is a tree containing every vertex of G.

• A simple graph is connected if and only if it has a spanning tree.

• Applied in IP multitasking.

Spanning TreesSpanning Trees

a b c

e f g

d

a b c

e f g

d

a b c

e f g

d

A simple graph

Not Spanning tree

Spanning tree

4.5 SPANN

ING TREES

1. Find a spanning tree for the following graphs.

EXERCISE 4.5EXERCISE 4.5

b

a c

e d

a b

d c

b

c

a

d

e f

h g

e

d

ca

b

FIGURE 1 FIGURE 2

FIGURE 3 FIGURE 4

4.5 SPANN

ING TREES

EXERCISE 4.5 : EXTRAEXERCISE 4.5 : EXTRA

PAGE : 734, 735, 736 and 737

Rosen K.H., Discrete Mathematics & Its Applications, (Seventh Edition), McGraw-Hill, 2007.

4.64.6 MINIMUM SPANNING TREESMINIMUM SPANNING TREES

• Find minimum spanning tree using Prim’s algorithm

• Find minimum spanning tree using Kruskal’s algorithm

CHAPTER 4 TREES

4.6 MIN

IMU

M SPAN

NIN

G TREES

• A minimum spanning tree in a connected weighted graph is a spanning tree that has the smallest possible sum of weights of it edges.

• Two algorithms can be used:– Prim’s Algorithm

• Robert Prim, 1957– Kruskal’s Algorithm

• Joseph Kruskal, 1956

Minimum Spanning TreesMinimum Spanning Trees

4.6 MIN

IMU

M SPAN

NIN

G TREES

1. Chose an edge with the least weight.2. Include it in spanning tree, T.3. Select an edge of least weight that is incident with a vertex

of an edge in T.4. If it does not create a cycle (simple circuit) with the edges in

T, then include it in T; otherwise discard it.5. Repeat STEPS 3 and 4 until T contains n-1 edges.

• There may be more than one minimum spanning tree for a given connected weighted simple graph.

• If there are two edges with similar smallest weight, chose either one.

Prim’s AlgorithmPrim’s Algorithm

4.6 MIN

IMU

M SPAN

NIN

G TREES

EXAMPLE: Prim’s AlgorithmEXAMPLE: Prim’s AlgorithmCHOICE EDGE WEIGHT SPANNING TREE

1 {a, b} 7

2 {a, d} 10

3 {b, e} 11

4 {e, c} 12

db

c

a e

7

1310

16

13

12

12

FIGURE 1

10 11

b

a7

db

a7 10

15

db

a7 10

db

a7 10

e11

e11

c12

The minimum spanning tree is given by:

db

a7 10

e11

c12

Total weight = 40

4.6 MIN

IMU

M SPAN

NIN

G TREES

1. Arrange the edges in G in increasing order. 2. Chose an edge with the minimum weight.3. Include it in spanning tree, T.4. Add an edge of least weight to T.5. If it does not create a cycle (simple circuit) with the edges in

T, then include it in T; otherwise discard it.6. Repeat STEPS 4 and 5 until T contains n-1 edges.

• There may be more than one minimum spanning tree for a given connected weighted simple graph.

• If there are two edges with similar smallest weight, chose either one.

Kruskal’s AlgorithmKruskal’s Algorithm

4.6 MIN

IMU

M SPAN

NIN

G TREES

EXAMPLE: Kruskal’s AlgorithmEXAMPLE: Kruskal’s AlgorithmCHOICE EDGE WEIGHT SPANNING TREE

1 {a, b} 7

2 {b, d} 10

3 {b, e} 11

4 {e, c} 12

db

c

a e

7

1310

16

13

12

12

FIGURE 1

10 11

b

a7

b

a7

10

15

b

a7

10

b

a7

10

e11

e11

c

12

The minimum spanning tree is given by:

b

a7

10

e11

c

12

Total weight = 40

d

d

d

d

4.6 MIN

IMU

M SPAN

NIN

G TREES

1. Construct a minimum spanning tree for each of the following connected weighted graphs using Prim’s and Kruskal’s algorithm.

EXERCISE 4.6EXERCISE 4.6

68

3

6

8

7

5 7

d

b f

c

a e2

2

3

5

11

3

2

FIGURE 1 FIGURE 2

1

18

b

c

a

d

ef

h g

1

1

1

1

4.6 MIN

IMU

M SPAN

NIN

G TREES

EXERCISE 4.6 : EXTRAEXERCISE 4.6 : EXTRA

PAGE : 742 and 743

Rosen K.H., Discrete Mathematics & Its Applications, (Seventh Edition), McGraw-Hill, 2007.

CHAPTER 4 TREES

SUMMARYSUMMARY

• Tree is a connected undirected graph with no simple circuits

• Trees have been employed to solve problems in a wide variety of disciplines.