computer notes - data structures - 15

8/3/2019 Computer Notes - Data Structures - 15

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Class No.15

Data Structures

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Level-order Traversal

There is yet another way of traversing abinary tree that is not related to recursivetraversal procedures discussed previously.

In level-order traversal, we visit the nodesat each level before proceeding to the nextlevel.

At each level, we visit the nodes in a left-to-right order.

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Level-order: 14 4 15 3 9 18 7 16 20 5 17

14

4

9

7

3

5

15

18

16 20

17

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How do we do level-order traversal?

Surprisingly, if we use a queue instead ofa stack, we can visit the nodes in level-order.

Here is the code for level-order traversal:



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Level-order Traversalvoid levelorder(TreeNode* treeNode)

{

Queue q;

if( treeNode == NULL ) return;

q.enqueue( treeNode);

while( !q.empty() )

{

treeNode = q.dequeue();

cout getInfo()) getLeft() != NULL )

q.enqueue( treeNode->getLeft());

if(treeNode->getRight() != NULL )q.enqueue( treeNode->getRight());

}

cout


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{

Queue q;



while( !q.empty() )

{





}

cout


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{

Queue q;



while( !q.empty() )

{





}

cout


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{

Queue q;



while( !q.empty() )

{





}

cout


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{

Queue q;



while( !q.empty() )

{





}

cout


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{

Queue q;



while( !q.empty() )

{





}

cout


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{

Queue q;



while( !q.empty() )

{





}

cout


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{

Queue q;



while( !q.empty() )

{





}

cout


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Queue: 14

Output:

14

4

9

7

3

5

15

18

16 20

17



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Queue: 4 15

Output: 14

14

4

9

7

3

5

15

18

16 20

17



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Queue: 15 3 9

Output: 14 4

14

4

9

7

3

5

15

18

16 20

17



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Queue: 3 9 18

Output: 14 4 15

14

4

9

7

3

5

15

18

16 20

17



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Queue: 9 18

Output: 14 4 15 3

14

4

9

7

3

5

15

18

16 20

17



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Queue: 18 7

Output: 14 4 15 3 9

14

4

9

7

3

5

15

18

16 20

17



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Queue: 7 16 20

Output: 14 4 15 3 9 18

14

4

9

7

3

5

15

18

16 20

17



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Queue: 16 20 5

Output: 14 4 15 3 9 18 7

14

4

9

7

3

5

15

18

16 20

17



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Queue: 20 5 17

Output: 14 4 15 3 9 18 7 16

14

4

9

7

3

5

15

18

16 20

17



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Queue: 5 17

Output: 14 4 15 3 9 18 7 16 20

14

4

9

7

3

5

15

18

16 20

17



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Queue: 17

Output: 14 4 15 3 9 18 7 16 20 5

14

4

9

7

3

5

15

18

16 20

17



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Queue:

Output: 14 4 15 3 9 18 7 16 20 5 17

14

4

9

7

3

5

15

18

16 20

17



25/55

Storing other Type of Data

The examples of binary trees so far havebeen storing integer data in the tree node.

This is surely not a requirement. Any typeof data can be stored in a tree node.

Here, for example, is the C++ code tobuild a tree with character strings.



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Binary Search Tree with Strings

void wordTree()

{

TreeNode* root = new TreeNode();

static char* word[] = "babble", "fable", "jacket",

"backup", "eagle","daily","gain","bandit","abandon",

"abash","accuse","economy","adhere","advise","cease",

"debunk","feeder","genius","fetch","chain", NULL};

root->setInfo( word[0] );

for(i=1; word[i]; i++ )

insert(root, word[i] );

inorder( root ); cout


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void wordTree()

{











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void wordTree()

{











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void wordTree()

{











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void wordTree()

{











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void insert(TreeNode* root, char* info)

{

TreeNode* node = new TreeNode(info);

TreeNode *p, *q;

p = q = root;

while( strcmp(info, p->getInfo()) != 0 && q != NULL )

{

p = q;

if( strcmp(info, p->getInfo()) < 0 )

q = p->getLeft();

else

q = p->getRight();

}



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{


TreeNode *p, *q;

p = q = root;


{

p = q;


q = p->getLeft();

else

q = p->getRight();

}



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{


TreeNode *p, *q;

p = q = root;


{

p = q;


q = p->getLeft();

else

q = p->getRight();

}



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{


TreeNode *p, *q;

p = q = root;


{

p = q;


q = p->getLeft();

else

q = p->getRight();

}



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{


TreeNode *p, *q;

p = q = root;


{

p = q;


q = p->getLeft();

else

q = p->getRight();

}



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if( strcmp(info, p->getInfo()) == 0 ){

cout setRight( node );

}



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}



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}



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Binary Search Tree with StringsOutput:

abandonabash

accuse

adhere

advise

babble

backup

banditcease

chain

daily

debunk

eagle

economy

fablefeeder

fetch

gain

genius

jacket



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Binary Search Tree with Stringsabandon

abashaccuse

adhere

advise

babblebackup

banditcease

chain

dailydebunk

eagle

economy

fablefeederfetch

gain

genius

jacket

Notice that the words are sorted inincreasing order when we traversedthe tree in inorder manner.



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abashaccuse

adhere

advise

babblebackup

banditcease

chain

dailydebunk

eagle

economy

fablefeederfetch

gain

genius

jacket


This should not come as a surprise ifyou consider how we built the BST.



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abashaccuse

adhere

advise

babblebackup

banditcease

chain

dailydebunk

eagle

economy

fablefeederfetch

gain

genius

jacket



For a given node, values less than theinfo in the node were all in the leftsubtree and values greater or equalwere in the right.



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abashaccuse

adhere

advise

babblebackup

banditcease

chain

dailydebunk

eagle

economy

fablefeederfetch

gain

genius

jacket




Inorder prints the left subtree, then thenode finally the right subtree.



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abashaccuse

adhere

advise

babblebackup

banditcease

chain

dailydebunk

eagle

economy

fablefeederfetch

gain

genius

jacket





Building a BST and doing an inordertraversal leads to a sorting algorithm.

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abashaccuse

adhere

advise

babblebackup

banditcease

chain

dailydebunk

eagle

economy

fablefeederfetch

gain

genius

jacket





Building a BST and doing an inordertraversal leads to a sorting algorithm.

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46/55

Deleting a node in BST

As is common with many data structures,the hardest operation is deletion.

Once we have found the node to bedeleted, we need to consider severalpossibilities.

If the node is a leaf, it can be deleted

immediately.



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If the node has one child, the node can bedeleted after its parent adjusts a pointer tobypass the node and connect to inordersuccessor.

6

2

4

3

1

8



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The inorder traversal order has to bemaintained after the delete.

6

2

4

3

1

8

6

2

4

3

1

8



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The inorder traversal order has to bemaintained after the delete.

6

2

4

3

1

8

6

2

4

3

1

8

6

2

31

8



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The complicated case is when the node tobe deleted has both left and right subtrees.

The strategy is to replace the data of this

node with the smallest data of the rightsubtree and recursively delete that node.



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Delete(2): locate inorder successor

6

2

5

3

1

8

4Inorder

successor



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Delete(2): locate inorder successor

6

2

5

3

1

8

4Inorder

successor

Inorder successor will be the left-mostnode in the right subtree of 2.

The inorder successor will not have a leftchild because if it did, that child would bethe left-most node.



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Delete(2): copy data from inorder successor

6

2

5

3

1

8

4

63

5

3

1

8

4



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Delete(2): remove the inorder successor

6

2

5

3

1

8

4

63

5

3

1

8

4

63

5

3

1

8

4



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Delete(2)

63

5

4

1

8

63

5

3

1

8

4
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computer notes - data structures - 15

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