Section 10.1 Introduction to Trees These class notes are based on material from our textbook,...

Section 10.1Introduction to Trees

These class notes are based on material from our textbook, Discrete Mathematics and Its Applications, 6th ed., by Kenneth H. Rosen, published by McGraw Hill, Boston, MA, 2006. They are intended for classroom use only and are not a substitute for reading the textbook.

Chapter 10

Tree

• A tree is a connected undirected graph with– No simple circuits– No multiple edges– No loops

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

Which graphs are trees?

a b

c d

e f

a b

c

d

e f

a b

c d

e f

a b

c

d

e f

YES YES NO NO

Forest

• What if there are no simple circuits but the graph is not connected?

• Each of the connected components is a tree

• The collection is called a forest.

Rooted Tree

• Specify a vertex as root, then direct each edge away from the root. The resulting tree is called a rooted tree.

Root

Example

Roota b c

d e fa

b

cd e f

a

b

c

d

e

f

What if a different root is chosen?

A different rooted tree results.

Roota b c

d e f

a

b c

d e f

a

b c

d

e f

a

b

c

d

e

f

Tree Terminology

• If v is a vertex of 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.

• When u is the parent of v, v is called the child of u.

• If two vertices share the same parent, then they are called siblings.

Example

a

b c d

e f g h

i j k l m

Root

Example

a

b c d

e f g h

i j k l m

Siblings

Tree Terminology (Cont.)

• 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.

Example

a

b c d

e f g h

i j k l m

Ancestors of k

a

b c d

e f g h

i j k l m

Example

Descendants of d


• A vertex with no children is called a leaf.

• Vertices with children are called internal vertices.

Example

a

b c d

e f g h

i j k l m

Leaves

Example

a

b c d

e f g h

i j k l m

Internal vertices


• 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.

Example

a

b c d

e f g h

i j k l m

Subtree at dSubtree at b


• 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.

Example

• What is the arity of this tree?

• Is this a full m-ary tree?-------------------

• This is a 2-ary, or binary, tree.

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

Example

• What is the arity of this tree?

• Is this a full m-ary tree?

--------• This is a 3-ary tree.• Yes, this is a full 3-

ary tree, since every internal vertex has exactly 3 children.

Example

• What is the arity of this tree?• Is this a full m-ary tree?

-------

• This is a full 5-ary tree.

Example

• What is the arity of this tree?• Is this a full m-ary tree?

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

It is not a full-3-ary tree, since one internal node has only 2 children.

Ordered Rooted Tree

• An ordered rooted tree is one where the children of each internal vertex are ordered.

• In an ordered binary tree, if an internal vertex has two children, then they are called left child and right child.

• The subtree rooted at the left child of a vertex is called the left subtree and subtree rooted at the right child of a vertex is called the right subtree.

Example

Left child of d

a

b c

d e h i

f g j k l

m

Example

Right child of d

a

b c

d e h i

f g j k l

m

Example

a

b c

d e h i

f g j k l

mLeft subtree of c

Example

a

b c

d e h i

f g j k l

m Right subtree of c


• The level of a vertex v in a rooted tree is the length of the unique path from the root to this vertex.

• What is the level of the root? 0

• The height of a rooted tree is the maximum of the levels of the vertices.

Example

a

b c d

e f g h

i j k l m

Levels 0

1

2

3

Height = 3

Properties of Trees

• A tree with n vertices has n1 edges.

• An full m-ary tree with i internal vertices contains n = mi + 1 vertices.

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

Example

Is this tree balanced?

Tree Properties (Cont.)

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

• If an m-ary tree with l leaves is full and balanced, then its height is

h = logml

Homework Exercises

• Do Section 10.1 exercises # 1, 3, 5, 7, 9, 17, 19, 20, 27, 28

CSE 2813Discrete Structures

Chapter 10, Section 10.2Applications of Trees


Applications of Trees

• Binary Search Trees• Decision Trees• Prefix Codes (Huffman Coding)• Game Trees

Full and Complete Binary Trees

A full binary tree is a binary tree in which each node is either a leaf node or has degree 2 (i.e., has exactly 2 children).

A complete binary tree is a full binary tree in which all leaves have the same depth.

A nearly complete binary tree is completely filled on all levels except possibly the lowest, which is filled from the left up to a point.

Examples

Full binary tree: Complete binary tree:

Binary Trees

• What is the smallest height possible in a binary tree of 7 nodes? How many leaf nodes does it have?

height = 2

num. leaves = 4

Binary Trees

•What is the smallest height possible in a binary tree of 15 nodes? How many leaf nodes does it have?

height = 3

num. leaves = 8

Binary Trees

• What is the smallest height possible in a binary tree of 31 nodes? How many leaf nodes does it have?

height = 4

num. leaves = 16

Binary Trees

• What is the smallest height possible in a binary tree of (2n) - 1 nodes?

• The smallest height possible in a binary tree of (2n) - 1 nodes is n – 1.

• Example: a tree with 31 nodes has 25 – 1 nodes, so n = 5, and its height = (n – 1) = (5 – 1) = 4.

Binary Trees

• How many leaf nodes does a binary tree of (2n) - 1 nodes have?

• A tree with (2n) - 1 nodes has 2n-1 leaves

• Example: A tree with 31 nodes has 25 – 1 nodes, so n = 5, and this tree has 2n-1 = 25-1 = 24 = 16 leaves.

Binary Trees

Note the pattern here:In a completely filled binary tree

with (2n) – 1 nodes, half of the nodes (rounding up) will be leaves. That is, (2n) / 2 nodes will be leaf nodes. And we can rewrite (2n) / 2 as 2n-1.

Binary Trees

Lemma: For any h 1, a binary tree which has more

than 2h-1 leaf nodes must have a height greater than h – 1.Example:

If a binary tree has 17 leaf nodes, can it have a height of 4?

No; a complete binary tree of height 4 has only 16 leaf nodes. A binary tree with 17 leaves must have a height greater than 4.

Binary Search Trees

Decision Trees

Prefix Codes

Huffman Codes

Consider the problem of data compression. Suppose that we have a 100,000-character file that contains only 6 different characters, a - f. Some 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, ..., f = 101.

This requires 300,000 bits to store the file.

This slide and the next 53 are adapted from: Cormen, Leiserson, Rivest, and Stein, Introduction to Algorithms, 2nd edition, The MIT Press, McGraw-Hill, 2001.

Huffman Codes

Huffman coding 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; this requires that the bit-patterns for our other letters be 3 or 4 bits long.

Using Huffman coding, we can store the file in 2240,00 bits.

Huffman Codes

a 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

Since we are using a prefix code, we can simply concatenate our codewords together to produce our bitstring.

For example, the string abc can be represented as 0101100.

This is unambiguous. Why?

Huffman Codes

Only 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.

Huffman Codes

Huffman Coding, using a

“Greedy Algorithm”

Game Trees

CSE 2813Discrete Structures

Chapter 10, Section 10.3Tree Traversal


Universal Address Systems

To totally order the vertices of on orered rooted tree:

1. Label the root with the integer 0.

2. Label its k children (at level 1) from left to right with 1, 2, 3, …, k.

3. For each vertex v at level n with label A, label its kv children, from left to right, with A.1, A.2, A.3, …, A.kv.

This labeling is called the universal address system of the ordered rooted tree.

Universal Address Systems

Traversal Algorithms

• A traversal algorithm is a procedure for systematically visiting every vertex of an ordered rooted tree– An ordered rooted tree is a rooted tree where

the children of each internal vertex are ordered

• The three common traversals are: – Preorder traversal

– Inorder traversal

– Postorder traversal

Traversal

• Let T be an ordered rooted tree with root r.• Suppose T1, T2, …,Tn are the subtrees at r

from left to right in T.

r

T1 T2 Tn

Preorder Traversal

r

T1 T2 Tn

Step 1: Visit rStep 2: Visit T1 in preorderStep 3: Visit T2 in preorder

.

.

.Step n+1: Visit Tn in preorder

Preorder Traversal

Example

A R EY PM HJ Q T

A

R

EY

P

M

HJ

Q T

Tree:

Visiting sequence:

The Preorder Traversal of T

In which order does a preorder traversal visit the vertices in the ordered rooted tree T shown to the left?Preorder:Visit root, then visit subtrees left to right.

Inorder Traversal

Step 1: Visit T1 in inorderStep 2: Visit rStep 3: Visit T2 in inorder

.

.

.Step n+1: Visit Tn in inorder

r

T1 T2 Tn

Example

A R EY PM HJ Q T

A

R

EY

P

M

HJ

Q T

Tree:

Visiting sequence:

The Inorder Traversal of T

In which order does an inorder traversal visit the vertices in the ordered rooted tree T shown to the left?

Inorder:Visit leftmost tree, visit root, visit other subtrees left to right.

Postorder Traversal

Step 1: Visit T1 in postorderStep 2: Visit T2 in postorder

.

.

.Step n: Visit Tn in postorderStep n+1: Visit r

r

T1 T2 Tn

Example

A R EYP MHJ Q T

A

R

EY

P

M

HJ

Q T

Tree:

Visiting sequence:

The Postorder Traversal of TIn which order does a postorder traversal visit the vertices in the ordered rooted tree T shown to the left?

Postorder:Visit subtrees left to right, then visit root.

Representing Arithmetic Expressions

• Complicated arithmetic expressions can be represented by an ordered rooted tree– Internal vertices represent operators

– Leaves represent operands

• Build the tree bottom-up– Construct smaller subtrees

– Incorporate the smaller subtrees as part of larger subtrees

Example

(x+y)2 + (x-3)/(y+2)

+

x y

2

–

x 3

+

y 2

/

+

Infix Notation

+

– +

/

+ 2

x y x 3 y 2

• Traverse in inorder adding parentheses for each operation

x+ y( ) 2( )+ x – 3( ) / y+2( )( )( )

Prefix Notation (Polish Notation)

• Traverse in preorder:

x+ y 2+ x– 3/ y+ 2

+

– +

/

+ 2

x y x 3 y 2

Evaluating Prefix Notation

• In an prefix expression, a binary operator precedes its two operands

• The expression is evaluated right-left

• Look for the first operator from the right

• Evaluate the operator with the two operands immediately to its right

Example

+ / + 2 2 2 / – 3 2 + 1 0+ / + 2 2 2 / – 3 2 1+ / + 2 2 2 / 1 1+ / + 2 2 2 1+ / 4 2 1

+ 2 1

3

Postfix Notation (Reverse Polish)

• Traverse in postorder

x +y 2 +x –3 /y +2

+

– +

/

+ 2

x y x 3 y 2

• In an postfix expression, a binary operator follows its two operands

• The expression is evaluated left-right

• Look for the first operator from the left

• Evaluate the operator with the two operands immediately to its left

Evaluating Postfix Notation

Example

3

2 2 + 2 / 3 2 – 1 0 + / +

4 2 / 3 2 – 1 0 + / +

2 3 2 – 1 0 + / +

2 1 1 0 + / +

2 1 1 / +

2 1 +

Homework Exercises

• Do questions :

7, 9, 10, 12, 13, 15, 16, 17, 18, 23, 24

Conclusion

• In this chapter we have studied:– Trees– Applications of Trees– Tree Traversal

Section 10.1 Introduction to Trees These class notes are based on material from our textbook,...

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