18569_TREES [Read-Only] [Compatibility Mode].pdf

8/10/2019 18569_TREES [Read-Only] [Compatibility Mode].pdf

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PROBLEMS AND DIFFICU

GIVE US THE CHANCE

BECOME STRONGER A

BETTER.


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TREESTable of Contents1) Introduction to Trees

Tree TerminoloTrees Properties

2) Binary TreesComplete Binary Trees

Extended Binary Trees3) Memory Representation

Sequential

Linked4) Binary Tree Traversal

Pre-orderIn-orderPost-order

5) Traversal Algorithms using Stacks


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President-CEO

ProductionManager Man

Shippin

Supervis

Personnel

Manager

Warehouse

Supervisor

Purchasing

Supervisor

HIERARCHICAL TREE STRUC TURE


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A tree is a finite set of one or msuch that:

There is a specially designated n

Definition of Tree

.The remaining nodes are partitioneddisjoint sets T1, ..., Tn, where each of ta tree.

We call T1, ..., Tn the subtrees of the ro


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Tree Terminology


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Terminology

The degree of a node is the number ofof the node

The degree of A is 3; the degree of C is 1.

The node with degree 0 is a leaf or

.A node that has subtrees is the parentroots of the subtrees.

The roots of these subtrees are the chithe node.

Children of the same parent aresiblings.The ancestors of a node are all thealong the path from the root to the node.


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Store elements Hierarchically

The top element: root

,

has a parent

Each element has 0 or more

children


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Ancestors of u


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Descendants of u


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Level Number

Each node in a binary tree T is assigNumber.

The root R of the tree T is assigned a level

one more than the level number of its paren

The nodes with the same level numberbelong to the same generation.


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A

HG

FE


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Node B has height 3, Node B has De

Node E has height 2, Node E has D

Node K has height 1, Node K has De


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Exercise

Number of Nodes Degree of a node(shown by green number) Leaf node(terminal) Non terminal nodes 3

Parent Children Sibling Ancestor Descendant

Level of a node Height(depth) of a tree

2 1 3

2 0 0 1 0

0 0 0


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Exercise to do

A

B C

Property

Number of nodes

Height

Root Node

D

G

E F

IH

Leaves

Interior nodes

Number of levels

Ancestors of H

Descendants of BSiblings of E

Degree of node A


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Binary Trees

ABinary Tree T is defined as a finite set of enodes , such that:

a)Tis empty (called null tree or empty tree) o

,and the remaining nodes of T form an o

disjoint binary treeT1(left subtree) andT2(

[This definition of the binary tree is recursive

Any node in Binary tree T has either 0,1 or 2A special class of trees: max degree for each no


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Example

A

B

CE

IM

HL

DF GK

Left Subtree and Right Subtree ?


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Difference between Binary tree and a Gen

A binary tree can be empty whereas a tr

empty.

Each element in binary tree has at most t

whereas each element in a tree can have an

sub trees

The sub trees of each element in a bin

ordered. That is we can distinguish betw

right sub trees. The sub trees in a tree are u


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Various Binary TreesSkewed Binary Tree: If every node in a tree has only one childnode), then we call such treesSkew Trees. If every node has only

call them as Left skew trees. Similarly, if every node has only ri

call them right skew trees.

A

B

C

D

Skewed Tree


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Complete Binary Tree

The tree T is said to be

complete if all its levels,

except possibly the last, have

e max mum num er opossible nodes, and if all the

nodes at the last level appear

as far left as possible.E

I

D

H


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Extended Binary Tree

A binary tree T is said to be a 2-tree or an extended binary t

N has either 0 or 2 children.

The nodes with 2 children are called internal nodes.

The nodes with 0 children are called external nodes.

The external and internal nodes are distinguished diagramm

circles for internal nodes and squares for external nodes.

A

B

D

B C

G

DF

Fig: Binary Tree T Fig: Extended 2-tree


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Maximum Number of Nodes in Bina

The maximum number of nodes onbinary tree is 2i, i>=0.

The maximum number of nodes in tree of depth kis 2k-1, k>=1.


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Full Binary Tree V/S Complete Bin

A full binary tree of depth k is a binary trehavin2k-1nodes,k>=0.

A binary tree !ith n nodes and "ompleteiffits nodes "orrespond to the nodfrom 1 tonin the full binary tree of depthk.

A

B C

GE

I

D

H

F

B

ED

JIH

Full binaClete binary tree


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Complete Binary Tre

#f a "omplete binary tree !ithnnodes

(depth=log n + 1)

is represented se$uentially,

t en or any no e ! t n ex , 1%= %=nparent'i( is ati)2*ifi+=1. #fi=1,iis ahas no parent.

leftChild'i( is at 2i if 2i%=n. #f 2i>n,

left "hild.rightChild'i( is at 2i1 if 2i1%=nthenihas no riht "hild.


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Representing Binary Trees in M

Binary trees are represented using two w

Sequential Representation (using Single Array)

[ The main Requirement of any representation of T is th

direct access to the root R of T and given any node N ofdirect access to the children of N. ]


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Advantages of Sequential Representation

%'

1. No need to store left and right pointers in the nodes2. Direct access to nodes: to get to node k, access A[k

TR

1 45

2 22

3 77

4 11

5 30

6

7 90

22 ((

11 30 )0

**2'1'

8

9 15

10 25

11

12

13

14 88

15

16

.

.

.

29


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It can be seen that a sequential representatio

tree requires numbering of nodes; starting w

level 1, then on level 2 and so on. Th

numbered from left to right .

It is an ideal case for representation of a co

tree and in this case no space is wasted.

o er nary rees, mos o e space remaAs can be seen in the figure, we require 1

array even though the tree has only 9 nodes.

for successors of the terminal nodes are

would actually require 29 locations insteasequential representation is usually ineff

binary tree is complete or nearly complete


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Sequential Representation is usually inefficient unless, the Binary trT is complete or nearly complete.

A

B

++

C

++

,1-

,2-

,3-

,%-

,'-

,.-

A

B

A

(1) waste space(2) insertion!e"etio

pro#"e$

++

++

D

++

/

E

(,*-

,)-

/

,1.-

E

C

DB

E

I

D

H


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Linked Representati

Tree T will be maintained in memory by me

Representation which uses three parallel

LEFT and RIGHT, and a pointer variable ROO

1. INFO[K] contains the data at the node

. con a ns e oca on o ethe node N

3. RIGHT[K] contains the location of th

of node N.

[Where K is the location ]

le!t


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Linked Representati

ointer variable //T "ontains thethe root of T.

#f the tree T ids empty, then //T

.#f any subtree of the tree is empt"orrespondin pointer !ill "onta

value.

"atale!t


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ExerciseConstruct a Binary tree using Linked Repr

INFO LEFT

1 K 0

2 C 3

3 G 0

4 14

5 A 10

6 H 17

5

Root

Avail

7 L 0

8 9

9 4

10 B 18

11 19

12 F 0

13 E 12

14 15

15 1616 11

17 J 7

18 D 0

19 20

20 0


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Binary Tree Travers

There are three standard !ays of tbinary tree T !ith root .

These three alorithms, "alled&

#norder

ostorder


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Preorder

1. Process the root R.

2. Traverse the left subtree of R in preorder.

3. Traverse the right subtree of R in preorder.

Inorder

1. Traverse the left subtree of R in inorder.


3. Traverse the right subtree of R in inorder.

Postorder

1. Traverse the left subtree of R in postorder.

2. Traverse the right subtree of R in postoder.



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Preorder

a

b c

a b c

1. Process the root R.2. Traverse the left subtree of R in preorder.

3. Traverse the right subtree of R in preorder.


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Preorder Example

a

b c

d e

g h i j

a b d g h e i c f j


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Preorder of Expression Tr

+

e

*

/

+

a b

-

c d

Gives prefix form of expression !

/ * + a b - c d + e f


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Inorder Example

a

b c

b a c

1. Traverse the left subtree of R in inorder.

2. Process the root R.3. Traverse the right subtree of R in inorder.


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Inorder of Expression Tre

+

e

*

/

+

a b

-

c d

Gives Infix form of expression !

ea + b * c d / +-


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Postorder Example

a

b c

b c a

1. Traverse the left subtree of R in postorder.

2. Traverse the right subtree of R in postoder.



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Postorder Example

a

b c

d e

g h i j

g h d i e b j f c a


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Postorder of Expression T

+

e

*

/

Gives postfix form of expression!

a b + c d - * e f + /

+

a b

-

c d


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Illustrations for Travers

Assume: visiting a node

is printing its label

Preorder:

3

85

1 3 5 4 6 7 8 9 10 11 12

Inorder:

4 5 6 3 1 8 7 9 11 10 12

Postorder:4 6 5 3 8 11 12 10 9 7 1

1

4 6


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Illustrations for Traversals (C

Assume: visiting a node

is printing its data

Preorder: 15 8 2 6 3 7

15

8

11 10 12 14 20 27 22 30

Inorder: 2 3 6 7 8 10 11

12 14 15 20 22 27 30

Postorder: 3 7 6 2 10 1412 11 8 22 30 27 20 15

6

3 7

10 12


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Arithmetic Expression Using Bina

E

#rer 4#re!i5

A

4

D

C

B

nrA 4 B in!i5 e

#t

A B 4 #t!i


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Some Examples

preorder = ab a

b

a

inorder = ab b

a

a

postorder = ab b

a

b

level order = ab a

b

a


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Binary Tree Constructi

Can you construct the binary tre

two traversal sequences?

Depends on which two sequence

given.


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Construct a Binary Tree from g

Inorder and Preorder Sequen

Inorder Sequence : D B E A F C

Preorder Sequence : A B D E C F


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Problem: Create a tree from the given traversals

preorder: F A E K C D H G B

inorder: E A C K F H D B GSolution: The tree is drawn from the root as follow

(a) The root of tree is obtained by choosing the firpreorder. Thus F is the root of the proposed tre

b The left child of the tree is obtained as follows(a) Use the inorder traversal to find the nodes to the le

root node selected from preorder. All nodes to the this case F) in inorder form the left subtree of the rC K )

(b) All nodes to the right of root node (in this case F ) right subtree of the root (H D B G)

(c) Follow the above procedure again to find the subsetheir subtrees on left and right.


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F is the root Nodes on left subtree( left of F):E A C K (fr

Nodes on right subtree(right of F):H D B G(fr

The root of left subtree:

From preorder: A E K C , Thus the root of left subtree is A D H G B , Thus the root of right subtree is

Creating left subtree first:

From inorder: elements of left subtree of A are: E (root of

elements of right subtree of A are: C K (roo

As K is to the left of C in preorder

A

F

D

KE

C


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Creating the right subtree of F

The root node is D From Inorder, the nodes on the left of D are: H (left ro

the nodes on the right of D are: B G (ri

Thus the tree is:

F

A D

E K HG

C B


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Inorder and Preorder

Preorder = a b d g h e i c f j

a

gdhbei fjc

b is the next root; gdh are in the lsubtree; ei are in the right subtree

a

gdh

fjcb

ei


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Inorder And Preorder

reorder = a b d h e i c f

a

gdh

fjcb

ei

d is the next root; g is in the lef

h is in the right subtree.a

g

fjcb

eid

h


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Preorder Traversal (using stackAlgorithm 9: PREORD(INFO, LEFT, RIGHT, ROOT)

The algorithm does a preorder traversal of Binary tree T, applying aneach of its nodes. An array STACK is used to temporarily hold the a

1. Set TOP:= 1, STACK[1]:= NULL and PTR := ROOT.

2. Repeat Steps 3 to 5 while PTR != NULL:

3. Apply PROCESS to INFO[PTR].

4. [Right Child?]

If RIGHT PTR != NULL thenSet TOP := TOP+1, and STACK[TOP]:= RIGHT[PTR].

[end of If structure]

5. [Left Child?]

If LEFT[PTR]!= NULL then

Set PTR:= LEFT[PTR].

ElseSet PTR:= STACK[TOP] TOP:= TOP-1.

[End of If structure]

[End of step 2 Loop]

6. Exit.


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6. Proceed down the left most path rooted at PTR= K as foll

(vi) Process K. (There is no right child)

No other node is processed, since K has no left child.7. [Backtracking] Pop C from Stack, and set PTR := C. This l

STACK:

Since PTR NULL, return to Step (a) of the algorithm.

8. Proceed down the left most path rooted at PTR = C as follo

(vii) Process C and push its right child F onto STACK:

STACK:

F

Preorder Traversal Algorithm (C

(viii) Process E. (There is no right child)

9. [Backtracking] Pop F from STACK, and set PTR :=F, This

STACK:

10. Proceed down the left-most path rooted at PTR :=F as fo

(ix)Process F.

No other node is processed, since F has no left child.

11. [Backtracking] Pop the top element NULL from STACK,

NULL.

Since PTR=NULL, the algorithm is completed.

As seen from the steps 2, 4, 6, 8 and 10, the nodes are proces

B, D, G, H, K, C, E, F. This is required preorder traversal.


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Inorder Traversal (using stack)

Algorithm 9: INORD(INFO, LEFT, RIGHT, ROOT)The algorithm does a inorder traversal of Binary tree T, applying an op

each of its nodes. An array STACK is used to temporarily hold the add


2. Repeat while PTR != NULL:

a) Set TOP:=TOP+1, STACK[TOP]:=PTR.

b Set PTR:= LEFT[PTR].[End of loop]

3. Set PTR:=STACK[TOP], TOP:=TOP-1.

4. Repeat Steps 5 to 7 while PTR != NULL:

5. Apply PROCESS to INFO[PTR].

6. If RIGHT[PTR]!= NULL, then:

a) Set PTR:=RIGHT[PTR]

b) Go to Step 2.[End of If structure]

7. Set PTR:=STACK[TOP] and TOP:=TOP-1.

[End of step 4 Loop.]

8. Exit


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Postorder Traversal (using stac

Algorithm 9: POSTORD(INFO, LEFT, RIGHT, ROOT)

The algorithm does a postorder traversal of Binary tree T, applying an

each of its nodes. An array STACK is used to temporarily hold the add


2. Repeat Step 3 to 5 while PTR!=NULL:

3. Set TOP := TOP+1, STACK[Top]:=PTR.

4. If RIGHT[PTR]!=NULL, then:

Set TOP := TOP+1, Stack[Top]:=-RIGHT[PTR].

End of If structure5. Set PTR:=LEFT[PTR]

[End of step 3 Loop]

6. Set PTR:=STACK[Top], TOP := TOP-1.

7. Repeat while PTR>0:

a)Apply PROCESS to INFO[PTR].

b)PTR:=STACK[Top], TOP := TOP-1.

[End of Loop]8. If PTR


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Similar and Copy Tree

A

B

E

F

A

B

C D GH C D

Tree T1 Tree T3Tree T2

1) Tree T1,T2 and T3 is said as Similar Tree because Str

trees is same.2) Tree T1 and T3 is said as Copy Tree because along w

content(data) part of both trees are same.3) Tree T1 and Tree T4 are neither similar nor Copy tree.


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Exercise

Q1: The post order traversal of a binary tree is 452631. Find outorder traversala. 126345b. 142653

c. 124536d. 124356

Q2: The inorder and preorder traversals of a binary tree are d h and a b d h e c f g I j respectively. The Post order traversal is

a) h d e b f i j g c ab) d h e b i j g f c ac) e h d b f i j g c a

D) h e d b f i j g a c


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