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Dr. Ahmad R. Hadaegh

A.R. Hadaegh Cali for nia State University San Marcos (CSUSM) Page 1

AVL Trees

and

Heaps

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AVL Trees

So far balancing the tree was done globally

Basically every node was involved in the balance operation

Tree balancing can however be done locally if only aportion of the tree is affected. An example of this is called

AVL tree

AVL tree was proposed by Adelson, Velskii and Landis

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AVL Trees

An AVL tree is one in which the height of the left and rightsub-trees of every node differ by at most one

For example consider the following tree

19| 08| +1

15| -1

12| 0

There are two values in each node: one is

the actual value of the node and the other is

called balancing factorof the node.

The balancing factor of every node in an AVL tree is either 0, +1, or -1.

For each node, P, the

balancing factor of P = height of right sub-tree of Pheight of left sub-tree of P

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AVL Trees -- Continue

If a new node is inserted to an AVL tree or a node is deleted

from an AVL tree, the balancing factor of some nodes

changes. In order to make the new modified tree an AVL

tree again we may have to rotate some nodes

Insertion or deletion may cause the balance factor of some

nodes to be +2 or -2. These nodes have to be tracked and

based on appropriate algorithms they should be properly

rotated to right/left to make the tree a balanced AVL tree

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Inserting into an AVL Tree

Algorithm:

Insert the new node, node N, into the tree the same way youinsert N into a binary search tree

Follow the path from node N to the root. This is your insertion

path

Insertion of node N may change the balancing factors of somenodes in the insertion path

After inserting node N, start updating the balancing factors ofthe nodes in the insertion path until you reach the first nodewith balancing factor of + 2 or -2. If such a node does not existin the insertion path, the tree is already balanced; otherwisemark that node, node P and go to the next step

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R = P -> Right

L = P -> Left

If the new node that was just inserted, Node N, is in theright sub-tree of R

Left Rotate R around P

else if Node N is in the left sub-tree of R

C = R ->Left

Right rotate C around R Left rotate C around P

else if Node N, is in the left sub-tree of L

Right Rotate L around P

else if Node N is in the right sub-tree of L C = L ->Right

Left rotate C around L

Right rotate C around P

Case1

Case 2

Case 3

Case 4

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7/36Dr. Ahmad R. HadaeghA.R. Hadaegh Cali for nia State University San Marcos (CSUSM) Page 7

L R

P

C C

.

.

.

Right rotateof C around

Rthen,

left rotate ofC around P

Left rotate

of R

around P

Right

rotate of L

around P

Left rotate ofC around L

then,right rotate of

C around P

Suppose node P is the first

node on the insertion path

in which its balance factor

is changed to +2 or -2

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Example of Case 1:

23| 0

19| +18 | 0

15| 0

12 | 05 | 0

23| ?

19| ?8 | 0

15| ?

12 | 05 | 0

30| 0

Step 1

Step 2

Original Tree

Inserting the

node 30 into the

tree

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23| +1

19| +28 | 0

15| ?

12 | 05 | 0

30| 0

Stop HereP

R

23| 0

19| 0

8 | 0

15| 0

12 | 05 | 0 30| 0

R

P

Step 3

Step 4

Following the

insertion path to

find node P withbalancing factor

+2 or -2

R is rotated

around P tobalance the tree

(case 1)

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Example of Case 2:

23| 0

19| +18 | 0

15| 0

12 | 05 | 0

23| ?

19| ?8 | 0

15| ?

12 | 05 | 0

21| 0

Original Tree

Inserting the

node 21 into the

tree

Step 1

Step 2

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23| -1

19| +28 | 0

15| ?

12 | 05 | 0

21 | 0

Stop HereP

R

C

21| 0

19| 0

8 | 0

15| 0

12 | 05 | 0 23| 0

C

P R21| ?

19| ?8 | 0

15| ?

12 | 05 | 0

23| ?R

P

C

Following the

insertion path to


+2 or -2

Making a double rotations to balance the tree (case 2)

Step 3

Step 4

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Example of Case 3:

19| 08 | 0

15| -1

12 | 05 | 0

19| 08 | ?

15| ?

12 | 05 | ?

2| 0

Step 1

Step 2

Original Tree

Inserting the

node 2 into the

tree

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P

L

5 | -1

8| 0

12 | 02 | 0

P

Step 3

Step 4

Following the

insertion path to


+2 or -2

L is rotated

around P tobalance the tree

(case 3)

19| 08 | -1

15| -2

12 | 05 | -1

2| 0

Stop Here

15| 0

19| 0

L

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Example of Case 4:

19| 08 | -1

15| -1

5 | 0

19| 08 | ?

15| ?

5 | ?

6| 0

Step 1

Step 2

Original Tree

Inserting the

node 6 into the

tree

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Step 3

Step 4

Following the

insertion path to


+2 or -2

P

L

19| 08 | -2

15| ?

5 | +1

6 | 0

Stop

Here

C

PL

19| 06 | 0

15| -1

5 | 0 8 | 0

C

Making a double rotations to balance the tree (case 4)

P

L

19| 08 | ?

15| ?

6 | ?

5 | ?

C

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Inserting a series of numbers into a AVL Tree

We want to insert numbers 2, 8, 6, 10, 15, 4, 3, 20, 13, 18

2 | 0Insert 2

Insert 8

2 | +1

8 | 0

No rotation is required

because there is no

node with balance

factor of +2 or -2


because there is no

node with balance

factor of +2 or -2

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Insert 6

Insert 10

2 | +2

8 | -1

6 | 0

2 | +2

6 | +1

8 | 0

P

R

C

6 | 0

8 | 02 | 0

P

C

R

C

RP

Double Rotation is required

6 | +1

8 | +12 | 0

10 | 0


because there is no

node with balance

factor of +2 or -2

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Insert 15

Insert 4

6 | ?

8 | +22 | 0

10 |+1

15 | 0

P

R

6 | +1

10 | 02 | 0

15 |08 | 0

R

P

6 | 0

10 | 02 | +1

15 |08 | 04 | 0


because there is nonode with balance

factor of +2 or -2

Single rotation is required

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Insert 3

6 | ?

10 | 02 | +2

15 |08 | 04 | -1

3 | 0

P

R

C

6 | ?

10 | 02 | +2

15 |08 | 0

4 | 0

3 | +1

P

R

C

6 | 0

10 | 03 | 0

15 |08 | 02 | 0 4 | 0

PR

C

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Insert 20 6 | +1

10|+13 | 0

15|+18 | 02 | 0 4 | 0

20 |0

Insert 13 6 | +1

10|+13 | 0

15|08 | 02 | 0 4 | 0

20 |013 | 0


because there is no node

with balance factor of +2

or -2


because there is no node

with balance factor of +2

or -2

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Insert 186 | +1

10|+23 | 0

15|+18 | 02 | 0 4 | 0

20 |-113 | 0

18 | 0

RL

P

6 | +1

15|03 | 0

20|-110 | 02 | 0 4 | 0

8 |-1 13 | 0 18 | 0

Single rotation is required

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Deleting From an AVL Tree

This can be more time consuming than insertion. First we need toapply a delete method to get rid of the node

We can use the delete method that finds either the successor orpredecessor node to replace a node that has both left and rightchildren

After the node is deleted, the balance factors are updated from theparent of the deleted nodeup to the root. This is the delete path

For each node on the delete path whose balance factor becomes +2or -2, a single or a double rotations has to be performed to restorethe balance of the tree

IMPORTANT NOTE: The balancing of the tree DOES NOTSTOP after the first node P with balancing factor +2 or -2 found inthe delete path. You still need to continue searching for other nodeswith balancing factor +2 or -2

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Delete Algorithm -- AVL Tree

1. Start from the parent of the deleted node and move up the

tree (delete path)2. If you find a node with balance factor +2 or -2 then stop

and do the balancing as follows3. Mark the node P4. If the deleted node is in the left sub-tree of P

Q = P -> RightIf balance factor of Q is +1 or 0

rotate Q around Pelse // the balance factor is -1

R = Q -> Left

rotate R around Qrotate R around P

endif

5. else if the .

Case1

Case2

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Delete Algorithm -- AVL TreeCont.

5. else if the deleted node is in the right sub-tree of P

Q = P -> Leftif balance factor of Q is -1 or 0

rotate Q around P

else// the balance factor is +1

R = Q -> Right

rotate R around Q

rotate R around P

end if

6. Continue moving up the tree on the delete path. If youfind another node with balancing factor +2 or -2 go to

step 3; otherwise, you are done

Case3

Case4

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Example of Case 1:

25| 0

20| +15 | 0

10| +1

8 | 03 | 0

Step 1

Original Tree

Delete 10

Delete method:

Find successor of10 which is 15

15| 0

30| 023| 0

25| 0

20| +15 | 0

10| +1

8 | 03 | 0

Step 2

15| 0

30| 023| 0

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Dr. Ahmad R. HadaeghA.R. Hadaegh Cali for nia State University San Marcos (CSUSM) Page 26

Start from node

20 (the parent of

deleted node)

and find the firstnode with

balancing factor

+2 or -2

25| 0

20| +25 | 0

15| ?

8 | 03 | 0

Step 4

30| 023| 0

25| 0

20| ?5 | 0

15| ?

8 | 03 | 0 10| 0

30| 023| 0

Swap 10 and 15

And

Delete 10

P

Step 3

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Because thedeleted node wasin the left side of

P, we have

Q = P -> Right

Note: Balancefactor of Q is 0

Step 6

25| -1

23| 0

5 | 0

15| +1

8 | 03 | 0 30| 020|+1

Q

25| 0

20| +25 | 0

15| ?

8 | 03 | 0

30| 023| 0

P

Q

Because the

balancing factor of

Q was 0, we rotate

Q around P

Step 5

P

E l f C 2

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Example of Case 2:

25| -1

20| +15 | 0

10| +1

8 | 03 | 0

Step 1

Original Tree

Delete 10

Delete method:

Find successor of10 which is 15

15| 0

23| 0

25| -1

20| +15 | 0

10| +1

8 | 03 | 0

Step 2

15| 0

23| 0

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Start from node

20 (the parent of

deleted node)


balancing factor

+2 or -2

25| -1

20| +25 | 0

15| ?

8 | 03 | 0

Step 4

23| 0

25| -1

20| ?5 | 0

15| ?

8 | 03 | 0 10| 0

23| 0

Swap 10 and 15

And

Delete 10

P

Step 3

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Because the deletednode was in the leftside of P, we have

Q = P -> Right

Note: Balance factorof Q is -1, so

R = Q -> Left

Because the balancing factor of Q was -1, we do two rotations

25| -1

20| +25 | 0

15| ?

8 | 03 | 0

23| 0

P

Q

Step 5

R

Step 6

20| ?

25| ?

5 | 0

15| ?

8 | 03 | 0 23| ?

P

R

Q25| 0

23| 05 | 0

15| 0

8 | 03 | 0 20| 0

R

QP

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Start from node 5

(the parent of

deleted node)


balancing factor

+2 or -2Step 4

Swap 10 and 8

and

Delete 10Step 3

25| 0

20| 05 | ?

8| ?

10 | 03 | 0 15| 0

4 | 01 | 0

25| 0

20| 05 | -2

8| ?

3 | 0 15| 0

4 | 01 | 0

P

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Because thedeleted node wasin the right side

of P, we have

Q = P -> Left

Note: Balancefactor of Q is 0

Step 6

Because the

balancing factor of

Q was 0, we rotate

Q around P

Step 5

25| 0

20| 05 | -2

8| ?

3 | 0 15| 0

4 | 01 | 0

P

Q

25| 0

20| 0

5 | -1

8| -1

3 | +1

15| 0

4 | 0

1 | 0

P

Q

E l f C 4

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Example of Case 4:

25| 0

20| 05 | -1

10| -1

8 | 03 | +1

Step 1

Original Tree

Delete 8

Delete method:

Because node 8 isa leaf, it can be

simply deleted

15| 0

4 | 0

Step 2

25| 0

20| 05 | ?

10| ?

8 | 03 | +1 15| 0

4 | 0

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Start from node 5

(the parent of

deleted node)


balancing factor

+2 or -2

Step 4

Step 3

25| 0

20| 05 | -2

10| ?

3 | +1 15| 0

P

25| 0

20| 05 | -2

10| ?

3 | +1 15| 0

4 | 0

P

Q

Because the deletednode was in the

right side of P, wehave

Q = P -> Left

Note: Balance factorof Q is +1, so

R = Q -> Right

4 | 0

R

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25| 0

20| 05 | ?

10| ?

4 | ? 15| 0

3 | ?

P

Q

R

25| 0

20| 04 | 0

10| 0

3 | 0 15| 05 | 0PQ

R

Because the balancing factor of Q was +1, we do two rotations

Step 5

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