Topics in this Lecture

1
Topics in this Lecture

• Problem with BST-s
• Balanced Trees
• AVL Tree
• Rotations Single, Double

2
Problem of BST

• Recall the remove operation in the BST
• Remove replaces the node being removed by the
  minimum child in the right subtree of the node
  being removed
• This means that some node is getting removed and
  another node on the right subtree of the branch
  is replacing it

3
Left Heavy BST

• Repeated remove-s along the same branch of a BST
  results in the the right subtree losing nodes
  that replace a node somewhere to the left of that
  right subtree
• This results in BST-s becoming left heavy

4
Why are Left Heavy BST-s Bad?

• Because the average time for the tree operations
  deteriorates from O(log N) and for some specific
  cases it can become O(N2)
• The problem of left heavy BST-s can be solved
  using balanced trees

5
Balanced Trees

• Balanced trees require every node to be balanced
• An AVL (Adelson-Velskii and Landis named after
  the researches who proposed this data structure)
  tree is an example of a balanced tree
• AVL trees are identical to BST for operations
  such as insert, remove, find, etc.
• However, every time the number of nodes in the
  AVL tree changes (through an insert or delete)
  the tree needs to be balanced

6
AVL Tree

• To measure the balance at every node, a balance
  factor is stored at every node
• In an AVL tree every node stores the difference
  in height d between its left and right
  subtrees. This is the balance factor for an AVL
  tree
• If d 0 or d 1 or d -1 then the node is
  balanced

7
Examples of AVL Tree d
d-1
d1
d0
d0

• The d values above are for node A only
• Nodes B and C have d0 in all cases
• Nodes B and C can be single nodes(as shown) or
  subtrees

8
Allowable Values for d

• The value of d at a node can only get changed
  by an insert or a remove operation
• If d takes some other value because of an
  insert or a delete we have to adjust nodes so
  that all nodes in the AVL tree have d lt 1
• d stands for modulus of d. Modulus means we
  consider the value of d and not its sign
• So 0 0, 1 1, and -1 1

9
Where in the AVL Treemight d change?

• If a node is inserted or deleted under a node N,
  the d value can change at any node along the
  path starting from the root till node N

d-1
d-2 violation
insert node F
d-1
10
Example of insert in an AVL Tree
d-2 violation
d-1
insert node F
d-1
d-1

• All nodes where d values are not shown have d
  0
• After the insert the d values of nodes C and
  E become 1
• After the insert, the right subtree of A has
  height 3 while
• the left subtree has height 1 (only node B)
• So, the d value of node A is 2 after the
  insert.
• Since -2 2gt 1 therefore there is a
  violation of the balance factor at node A

11
Rotation in an AVL Tree

• To restore the d value of nodes where the d
  value has been violated, an operation called a
  rotation needs to be done
• Violation of the d value at a node is also
  called violation of balance or imbalance at that
  node
• There can be two types of rotation in an AVL tree
• single rotation
• double rotation

12
Conditions for Violation of Balance Condition

• For simplicity we will classify the operations at
  a node that might lead to an imbalance into four
  types
• So, an imbalance can be caused by one of the
  following four operations
• An insertion into the left subtree of the left
  child
• An insertion into the right subtree of the left
  child
• An insertion into the left subtree of the right
  child
• An insertion into the right subtree of the right
  child

13
Similarity between Insert and Remove on AVL Tree

• A remove is just the reverse of an insert.
• So the technique for removing an imbalance caused
  by an insert will apply in the same way for
  removing an imbalance caused by a remove.
• Hence, the cause of the imbalance from a remove
  can be classified into one of the four types on
  the last slide, except that, insertion into
  would be replaced by remove from.

14
Rotations in AVL Tree

• Single rotation is used to restore balance when
  conditions 1 or 4 (on slide 12) happen
• Double rotation is used to restore balance when
  conditions 2 and 3 (on slide 12) happen

15
Condition 1 Single Rotation Figure
d 1
16
Condition 1 Single Rotation

• In the figure on the left hand side of the
  previous slide, k2 is the left child of k1
• A node had been inserted in the left subtree X of
  the left child(k2) of k1
• This is similar to condition 1 mentioned on slide
  12
• As a result of the insertion, there is an
  imbalance at k1 as shown by its d-value2

17
Condition 1 Single Rotation (contd.)

• To restore balance, a single rotation operation
  is done as shown in the figure on slide 17
• After the single rotation notice that all nodes
  are balanced (i.e., they have d lt1)
• Also notice that k2 is the new parent node
• Verify that the BST property is preserved by the
  rotation

18
Condition 4 Single Rotation Figure
d 3-12
d2-11
k1
S. R.
1
k2
2
X
1
1
2
Z
Single Rotation
19
Condition 4 Single Rotation

• Condition 4 is just the mirror image of condition
  1
• In the figure on the left hand side of the
  previous slide, a node got inserted in the right
  subtree Z of the right child (k2) of node k1
• This corresponds to condition 4 on slide 14
• The node k1 got imbalanced due to this insertion
  and got a d-value 2

20
Condition 4 Single Rotation (contd.)

• To restore balance, once again a single rotation
  operation is done as shown in the figure on slide
  18
• After the single rotation notice that all nodes
  are balanced (i.e., they have d lt1)
• Also notice that k2 is the new parent node
• Verify once again that the BST property is
  preserved by the rotation

21
Condition 2 Double Rotation Figure
d 3-12
Double Rotation
1
d1-21
1
0.5
22
Condition 2 Double Rotation

• In the figure on the left hand side on the last
  slide, k1 is the left child of k3
• An insertion occurred into the right subtree of
  the left child (k1) of node k3
• This is condition 2 on slide 12

23
Condition 2 Double Rotation (contd.)

• The right subtree of k1 comprises node k2 and
  subtrees B and C.
• Before the insertion the right subtree of k1
  contained only node k2 (because the tree must
  have been balanced before the insertion).
• Due to the insertion only one of nodes B or C got
  inserted NOT BOTH

24
Condition 2 Double Rotation (contd.)

• B and C would be a single node subtree of height
  1
• To denote the average height, that is, either B
  or C is present (but not both) we show both B and
  C but give them a height 0.5
• Showing B and C both on the same figure with a
  height of 0.5 is just a short-cut used on paper
  to avoid drawing two separate diagrams one
  diagram for Bs insertion and another diagram for
  Cs insertion
• In real life (when the program runs), either B or
  C only gets inserted with a height of 1

25
Condition 2 Double Rotation (contd.)

• Due to the insertion an imbalance occurred at k3
• You should calculate the height of the left and
  right subtrees at k3 and verify that there is
  indeed an imbalance at k3
• While calculating the heights assume that only
  one of B or C is present with a height of 1
  instead of 0.5
• The imbalance at k3 is restored using the double
  rotation as shown on slide 23
• Verify once again that after the double rotation
  BST property is still valid and balance is
  restored at every node

26
Condition 3 Double Rotation

• Condition 3 is just the mirror image of condition
  2
• The corresponding figure is given as Figure 4.39
  on page 149 of the text book

27
AVL Tree Implementation Data Members

• The data structure for an AVL tree is identical
  to the BST
• Additionally, the balance factor needs to be
  stored at every node as an integer
• class AVLNode
• AVLNode lchild, rchild
• int balance
• int key

28
AVL Tree Operations

• The operations of BST such as insert, delete,
  find etc. can be used as is in AVL trees
• AVL trees require two additional methods one
  for single rotation and another for double
  rotation

29
Epilogue

• In this lecture we studied about an extension of
  BST-s called AVL trees
• AVL trees are height balanced
• Balance is maintained at each node by checking
  for an imbalance following an insert or remove
• If there is an imbalance, it is restored by doing
  a single or a double rotation, as need be, to
  restore the balance