Computer Notes - Data Structures - 23

8/3/2019 Computer Notes - Data Structures - 23

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Class No.23

Data Structures

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2/31

Expression Tree

The inner nodes contain operators while leafnodes contain operands.

a

c

+

b

g

*

+

+

d

*

*

e

f

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3/31

Expression Tree

The tree is binary because the operators arebinary.

a

c

+

b

g

*

+

+

d

*

*

e

f

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4/31

Expression Tree

This is not necessary. A unary operator (!, e.g.)will have only one subtree.

a

c

+

b

g

*

+

+

d

*

*

e

f



5/31

Expression Tree

Inorder traversal yields: a+b*c+d*e+f*g

a

c

+

b

g

*

+

+

d

*

*

e

f



6/31

Enforcing Parenthesis

void inorder(TreeNode* treeNode)

{

if( treeNode != NULL ){

cout getLeft());

cout


7/31

Expression Tree

Inorder : (a+(b*c))+(((d*e)+f)*g)

a

c

+

b

g

*

+

+

d

*

*

e

f



8/31

Expression Tree

Postorder traversal: a b c * + d e * f + g * +which is the postfix form.

a

c

+

b

g

*

+

+

d

*

*

e

f



9/31

Constructing Expression Tree

Algorithm to convert postfix expression into anexpression tree.

We already have an expression to convert aninfix expression to postfix.

Read a symbol from the postfix expression.

If symbol is an operand, put it in a one node treeand push it on a stack.

If symbol is an operator, pop two trees from thestack, form a new tree with operator as the rootand T1 and T2 as left and right subtrees andpush this tree on the stack.

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10/31


a b + c d e + * *

stack



11/31


a b + c d e + * *

ba

Stack is growing left to right

If symbol is anoperand, put it in

a one node treeand push it on astack.

top

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12/31


a b + c d e + * *

ba

Stack is growing left to right

+

If symbol is anoperator, pop two

trees from the stack,form a new tree withoperator as the rootand T1 and T2 as leftand right subtrees andpush this tree on the

stack.

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13/31


a b +c d e + * *

ba

+ dc e



14/31


a b +c d e+ * *

ba

+ c

ed

+



15/31


a b +c d e+* *

ba

+

c

ed

+

*



16/31


a b +c d e+**

ba

+

c

ed

+

*

*



17/31

Other Uses of Binary Trees

Huffman Encoding



18/31

Huffman Encoding

Data compression plays a significant role incomputer networks.

To transmit data to its destination faster, it isnecessary to either increase the data rate of the

transmission media or to simply send less data. Improvements with regard to the transmission

media has led to increase in the rate.

The other options is to send less data by means

of data compression. Compression methods are used for text, images,

voice and other types of data (space probes).



19/31

Huffman Encoding

Huffman code is method for the compression forstandard text documents.

It makes use of a binary tree to develop codes ofvarying lengths for the letters used in the original

message. Huffman code is also part of the JPEG image

compression scheme.

The algorithm was introduced by David Huffmanin 1952 as part of a course assignment at MIT.

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20/31

Huffman Encoding

To understand Huffman encoding, it isbest to use a simple example.

Encoding the 32-character phrase:"traversing threaded binary trees",

If we send the phrase as a message in anetwork using standard 8-bit ASCII codes,we would have to send 8*32= 256 bits.

Using the Huffman algorithm, we can sendthe message with only 116 bits.


21/31

Huffman Encoding

List all the letters used, including the "space"character, along with the frequency with whichthey occur in the message.

Consider each of these (character,frequency)

pairs to be nodes; they are actually leaf nodes,as we will see.

Pick the two nodes with the lowest frequency,and if there is a tie, pick randomly amongstthose with equal frequencies.


22/31

Huffman Encoding

Make a new node out of these two, and makethe two nodes its children.

This new node is assigned the sum of thefrequencies of its children.

Continue the process of combining the twonodes of lowest frequency until only one node,the root, remains.


23/31

Huffman Encoding

Original text:traversing threaded binary treessize: 33 characters (space and newline)

NL : 1SP : 3a : 3b : 1

d : 2e : 5g : 1h : 1

i : 2n : 2r : 5s : 2t : 3v : 1y : 1


24/31

Huffman Encoding

v

1

y

1

SP

3

r

5

h

1

e

5

g

1

b

1

NL

1

s

2

n

2

i

2

d

2

t3

a

3

2

2 is equal to sumof the frequencies ofthe two children nodes.


25/31

Huffman Encoding

v

1

y

1

SP

3

r

5

h

1

e

5

g

1

b

1

NL

1

s

2

n

2

i

2

d

2

t3

a

3

2 2

There a number of ways to combinenodes. We have chosen just one suchway.


26/31

Huffman Encoding

v

1

y

1

SP

3

r

5

h

1

e

5

g

1

b

1

NL

1

s

2

n

2

i

2

d

2

t3

a

3

2 2 2


27/31

Huffman Encoding

v

1

y

1

SP

3

r

5

h

1

e

5

g

1

b

1

NL

1

s

2

n

2

i

2

d

2

t3

a

3

2 2 2

4 4


28/31

Huffman Encoding

v

1

y

1

SP

3

r

5

h

1

e

5

g

1

b

1

NL

1

s

2

n

2

i

2

d

2

t3

a

3

2 2 2

5444

6


29/31

Huffman Encoding

v

1

y

1

SP

3

r

5

h

1

e

5

g

1

b

1

NL

1

s

2

n

2

i

2

d

2

t3

a

3

2 2 2

5444

86 9 10


30/31

Huffman Encoding

v

1

y

1

SP

3

r

5

h

1

e

5

g

1

b

1

NL

1

s

2

n

2

i

2

d

2

t3

a

3

2 2 2

5444

86

14

9

19

10


31/31

Huffman Encoding

v

1

y

1

SP

3

r

5

h

1

e

5

g

1

b

1

NL

1

s

2

n

2

i

2

d

2

t3

a

3

2 2 2

5444

86

14

9

19

10

33

Documents

Computer Notes - Data Structures - 23