# COP 3503 â€“ CSII â€“ Summer 2001 â€“ File by jcc96858

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```									  COP 3503H – Computer Science I – CLASS NOTES - DAY #26 Supplement

More on Tree Rotations

Single Rotations

Recall that the insertion of a new node into an AVL tree could occur in one of four
different ways (two of which are symmetric to each other):

1.   The new node is in the left subtree of the left child of a node.
2.   The new node is in the right subtree of the left child of a node.
3.   The new node is in the left subtree of the right child of a node.
4.   The new node is in the right subtree of the right child of a node.

Cases 1 and 4 are symmetric as are cases 2 and 3.

Whenever an insertion of case 1 or 4 occurs, the resulting tree will be unbalanced and a
single rotation can be used to rebalance the tree. Consider the following case 1
situation (insertion in left subtree of a left child):

P   -1

S     0                C   0

A                 B

initial AVL tree

Day 26 Supplement- 1
P       -2

S     -1                 C       0

B
A

tree immediately after insertion into subtree A

The new node has been inserted into the left subtree (A) of node S which is the left
child of node P. The tree is now unbalanced and is temporarily not an AVL tree. To
restore the balance in the tree and return it to the status of an AVL tree, a single
rotation of S about P is needed as shown below:

S        0

P
0

A

B                   C

Rebalanced tree is now an AVL tree

Day 26 Supplement- 2
Example – Case 1 Insertion

40                                               -1

30                        50                       0                           0

25              35                             0                  0

initial AVL tree                                  initial balance factors

40                                              -2

30                        50                      -1                          0

25               35                            -1                0

20                                         0

AVL tree after insertion of 20                    balance factors after insertion

Perform right rotation of 30 about 40 (40 is the unbalanced node).

40                                               30

30                        50                      25                          40

25              35                             20                          35            50

20

Day 26 Supplement- 3
The steps in the right rotation are:
1. Interchange nodes 30 and 40.
2. Interchange causes node 40 to move to right subtree of 30.
3. Former right subtree of 30 becomes left subtree of 40.
4. Left subtree of 30 is unchanged.
5. Right subtree of 40 is unchanged.

You should verify that the resulting tree is AVL balanced.

Now consider a case 4 situation (insertion in right subtree of a right child):

+1
P

0
C                           S

A                     B

initial AVL tree

P       +2

C                           S       +1

A

B

tree immediately after insertion into subtree B

Day 26 Supplement- 4
The new node has been inserted into the right subtree (B) of node S which is the right
child of node P. The tree is now unbalanced and is temporarily not an AVL tree. To
restore the balance in the tree and return it to the status of an AVL tree, a single
rotation of S about P is needed as shown below:

0
S

0
P

B

C             A

final AVL tree

Example – Case 4 Insertion

40                                             +1

30                  50                       0                   0

45          55                                 0         0

initial AVL tree                                initial balance factors

Day 26 Supplement- 5
40                                           +2

30               50                       0                    +1

45        55                                 0              +1

65                                                0

AVL tree after insertion of 65                balance factors after insertion

Perform left rotation of 50 about 40 (40 is the unbalanced node).

40                                                           50

30                50                                      40                      55

45         55                       30             45                      65
0

65
0

The steps in the left rotation are:
1. Interchange nodes 40 and 50.
2. Interchange causes node 40 to move to left subtree of 30.
3. Former left subtree of 50 becomes the right subtree of 40.
4. Left subtree of 40 is unchanged.
5. Right subtree of 50 is unchanged.

Day 26 Supplement- 6
Notice the symmetry of the two final AVL trees after the insertions and subsequent
rebalancing.

S                                                    S

P                              P

A                                                                    B
B           C                    C          A

case 1 final AVL tree                             case 4 final AVL tree

Double Rotations

Cases 2 and 3 when inserting into an AVL tree require a double rotation to rebalance
the tree. Case 2 occurs when the new node is in the left subtree of the right child of a
node. Case 3 occurs when the new node is in the right subtree of the right child of a
node. As before, cases 2 and 3 are symmetric.
Consider the following case 2 situation (insertion in left subtree of a right child):

G   0

P       +1

A                    S   0
D

B            C

initial AVL tree

Day 26 Supplement- 7
-2
G

P        +2

A                     S   -1
D

C

B

tree immediately after insertion into subtree B

The new node has been inserted into subtree B which is left subtree of node S which is
a right child of P. The tree is now unbalanced and is temporarily not an AVL tree. To
restore the balance in the tree and return it to the status of an AVL tree, a double
rotation of S about P followed by S about G is needed as shown below. While the
double rotation may seem confusing it is in reality two single rotations as shown
below:

G

P

A                    S
D

C

B

initial unbalanced tree

Day 26 Supplement- 8
G

S

P                 C
D

A

B

rotation of S about P (first rotation – tree still unbalanced)

S

P                   G

A                         C

B                       D

rotation of S about G (second rotation – tree becomes balanced)

Day 26 Supplement- 9
S

P           G

A                    C

D
B

final AVL tree (balanced)

Example – Case 2 Insertion (double left/right rotation)

40                                                  -1

30                  50                             0                   0

25              35                                 0             0

initial AVL tree                                 initial balance factors

40
-2

30                      50
+1                  0

25             35
0             -1

32
0

AVL tree after insertion of 32                   balance factors after insertion
Day 26 Supplement- 10
Perform double rotation: 1st do 35 about 30 followed by 35 about 40.

40                                                              40

30                   50                                           35                  50
step 1

25                  35                                             30

32                                               25              32

tree after insertion                                        tree after first rotation of 35 about 30

-2                                                          40

-2                   0                                           35                  50
step 2

0                                                                 30

0             0                                                25               32

balance factors after first rotation                        rotate 35 about 40

35
0

30                   40
0                  +1

25                   32             50
0                   0               0

tree after second rotation of 35 about 40                            final balance factors

Day 26 Supplement- 11
The steps in the double rotation are:
1. Interchange parent and grandparent of new node (nodes 35 and 30).
2. Interchange previous parent and great-grandparent (noded 35 and 40).

Example – Case 3 Insertion (double right/left rotation)

40                                              +1

30                  50                          0                   0

45             65                               0                0

initial AVL tree                              initial balance factors

40
+2

30                  50
0                   -1

45         65
+1               0

48
0

AVL tree after insertion of 48                balance factors after insertion

Day 26 Supplement- 12
Perform double rotation: 1st do 45 about 50 followed by 45 about 40.

40                                                            40

30                        50                                    30                    45
step 1

45                 65                                                       50

48                                                              48         65

tree after insertion                                            tree after first rotation of 45 about 50

+2                                                                 40

0                        +2                                        30                  45

step 2

0                                                           50

0              0                                             48             65

balance factors after first rotation                            rotate 45 about 40

45
0

40                           50
-1                  0

30                    48                65                  0                 0              0

tree after second rotation of 45 about 40                                final balance factors

Day 26 Supplement- 13
The steps in the double rotation are:
1. Interchange parent and grandparent of new node (nodes 35 and 30).
2. Interchange previous parent and great-grandparent (noded 35 and 40).

Day 26 Supplement- 14

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