AVL Trees Adelson-Velskii and Landis Binary Search Tree - Best Time All BST operations are O(d), where d is tree depth minimum d is for a binary tree with N nodes What is the best case tree? What is the worst case tree? So, best case running time of BST operations is O(log N) Binary Search Tree - Worst Time Worst case running time is O(N) What happens when you Insert elements in ascending order? Insert: 2, 4, 6, 8, 10, 12 into an empty BST Problem: Lack of “balance”: compare depths of left and right subtree Unbalanced degenerate tree Balanced and unbalanced BST Approaches to balancing trees Don't balance May end up with some nodes very deep Strict balance The tree must always be balanced perfectly Pretty good balance Only allow a little out of balance Adjust on access Self-adjusting Balancing Binary Search Trees Many algorithms exist for keeping binary search trees balanced Adelson-Velskii and Landis (AVL) trees (height-balanced trees) Splay trees and other self-adjusting trees B-trees and other multiway search trees Perfect Balance Want a complete tree after every operation tree is full except possibly in the lower right This is expensive For example, insert 2 in the tree on the left and then rebuild as a complete tree AVL - Good but not Perfect Balance AVL trees are height-balanced binary search trees Balance factor of a node height(left subtree) - height(right subtree) An AVL tree has balance factor calculated at every node For every node, heights of left and right subtree can differ by no more than 1 Store current heights in each node Height of an AVL Tree N(h) = minimum number of nodes in an AVL tree of height h. Basis N(0) = 1, N(1) = 2 Induction N(h) = N(h-1) + N(h-2) + 1 Solution (recall Fibonacci analysis) N(h) > h (  1.62) Height of an AVL Tree N(h) > h (  1.62) Suppose we have n nodes in an AVL tree of height h. n > N(h) (because N(h) was the minimum) n > h hence log n > h (relatively well balanced tree!!) h < 1.44 log2n (i.e., Find takes O(logn)) Node Heights Node Heights after Insert 7 Insert and Rotation in AVL Trees Insert operation may cause balance factor to become 2 or –2 for some node only nodes on the path from insertion point to root node have possibly changed in height So after the Insert, go back up to the root node by node, updating heights If a new balance factor is 2 or –2, adjust tree by rotation around the node Single Rotation in an AVL Tree Implementation Single Rotation Double Rotation Implement Double Rotation in two lines. Insertion in AVL Trees Insert at the leaf (as for all BST) only nodes on the path from insertion point to root node have possibly changed in height So after the Insert, go back up to the root node by node, updating heights If a new balance factor is 2 or –2, adjust tree by rotation around the node Insert in BST Insert in AVL trees Example of Insertions in an A

1. AVL Trees
Adelson-Velskii and Landis

2. Binary Search Tree - Best Time
• All BST operations are O(d), where d is
tree depth
• minimum d is for a binary tree
with N nodes
› What is the best case tree?
› What is the worst case tree?
• So, best case running time of BST
operations is O(log N)
2
d ëlog Nû 2 =

3. Binary Search Tree - Worst Time
• Worst case running time is O(N)
› What happens when you Insert elements in
3
ascending order?
• Insert: 2, 4, 6, 8, 10, 12 into an empty BST
› Problem: Lack of “balance”:
• compare depths of left and right subtree
› Unbalanced degenerate tree

4. Balanced and unbalanced BST
4
4
2 5
1 3
1
5
2
4
3
7
6
4
2 6
1 3 5 7
Is this “balanced”?

5. Approaches to balancing trees
• Don't balance
5
› May end up with some nodes very deep
• Strict balance
› The tree must always be balanced perfectly
• Pretty good balance
› Only allow a little out of balance
• Adjust on access
› Self-adjusting

6. 6
Balancing Binary Search Trees
• Many algorithms exist for keeping
binary search trees balanced
› Adelson-Velskii and Landis (AVL) trees
(height-balanced trees)
› Splay trees and other self-adjusting trees
› B-trees and other multiway search trees

7. 7
Perfect Balance
• Want a complete tree after every operation
› tree is full except possibly in the lower right
• This is expensive
› For example, insert 2 in the tree on the left and
then rebuild as a complete tree
Insert 2 &
complete tree
6
4 9
1 5 8
5
2 8
1 4 6 9

8. 8
AVL - Good but not Perfect Balance
• AVL trees are height-balanced binary
search trees
• Balance factor of a node
› height(left subtree) - height(right subtree)
• An AVL tree has balance factor
calculated at every node
› For every node, heights of left and right
subtree can differ by no more than 1
› Store current heights in each node

9. 9
Height of an AVL Tree
• N(h) = minimum number of nodes in an
AVL tree of height h.
• Basis
› N(0) = 1, N(1) = 2
• Induction
h
› N(h) = N(h-1) + N(h-2) + 1
• Solution (recall Fibonacci analysis)
› N(h) > fh (f » 1.62) h-1 h-2

10. 10
Height of an AVL Tree
• N(h) > fh (f » 1.62)
• Suppose we have n nodes in an AVL
tree of height h.
› n > N(h) (because N(h) was the minimum)
› n > fh hence logf n > h (relatively well
balanced tree!!)
› h < 1.44 log2n (i.e., Find takes O(logn))

11. 11
Node Heights
Tree A (AVL) Tree B (AVL)
height=2 BF=1-0=1
0 1
height of node = h
balance factor = hleft-hright
empty height = -1
1
2
0 0
0
6
4 9
1 5 8
1
0 0
6
4 9
1 5

12. 12
Node Heights after Insert 7
1 1
height of node = h
balance factor = hleft-hright
empty height = -1
2
3
0 0
1
6
4 9
1 5 8
0
2
0
6
1
4 9
1 5
07
07
balance factor
1-(-1) = 2
-1
Tree A (AVL) Tree B (not AVL)

13. Insert and Rotation in AVL Trees
• Insert operation may cause balance factor
to become 2 or –2 for some node
› only nodes on the path from insertion point to
root node have possibly changed in height
› So after the Insert, go back up to the root
node by node, updating heights
› If a new balance factor is 2 or –2, adjust tree
by rotation around the node
13

14. 14
Single Rotation in an AVL Tree
2
2
1
0 0
1
6
4 9
1 5 8
07
0
1
0
2
0
6
1
4
9
8
1 5
07

15. 15
Insertions in AVL Trees
Let the node that needs rebalancing be a.
There are 4 cases:
Outside Cases (require single rotation) :
1. Insertion into left subtree of left child of a.
2. Insertion into right subtree of right child of a.
Inside Cases (require double rotation) :
3. Insertion into right subtree of left child of a.
4. Insertion into left subtree of right child of a.
The rebalancing is performed through four
separate rotation algorithms.

16. 16
AVL Insertion: Outside Case
j
k
X Y
Z
Consider a valid
AVL subtree
h
h h

17. 17
AVL Insertion: Outside Case
j
k
X
Y
Z
Inserting into X
destroys the AVL
property at node j
h
h+1 h

18. 18
AVL Insertion: Outside Case
j
k
X
Y
Do a “right rotation”
Z
h
h+1 h

19. 19
Single right rotation
j
k
X
Y
Do a “right rotation”
Z
h
h+1 h

20. “Right rotation” done!
(“Left rotation” is mirror
symmetric)
20
Outside Case Completed
j
k
X Y Z
h
AVL property has been restored!
h+1
h

21. 21
AVL Insertion: Inside Case
j
k
X Y
Z
Consider a valid
AVL subtree
h
h h

22. Does “right rotation”
restore balance?
22
AVL Insertion: Inside Case
Inserting into Y
destroys the
AVL property
at node j
j
k
X
Y
Z
h
h h+1

23. 23
AVL Insertion: Inside Case
j
k
X
Y
“Right rotation”
does not restore
balance… now k is
out of balance
h+1 h
Z
h

24. 24
AVL Insertion: Inside Case
Consider the structure
of subtree Y… j
k
X
Y
Z
h
h h+1

25. 25
AVL Insertion: Inside Case
j
k
X
V
Z
W
i
Y = node i and
subtrees V and W
h
h h+1
h or h-1

26. 26
AVL Insertion: Inside Case
j
k
X
V
Z
W
i
We will do a left-right
“double rotation” . . .
h h or h-1
h

27. 27
Double rotation : first rotation
j
k
X V
Z
W
i
left rotation complete
h
h or h-1
h or h-1
h

28. Double rotation : second rotation
28
j
k
X V
Z
W
i
Now do a right rotation
h
h
h or h-1
h or h-1

29. Double rotation : second rotation
right rotation complete
Balance has been
restored
29
X
i
k j
V W Z
h h or h-1 h

30. 30
Implementation
balance (1,0,-1)
key
left right
No need to keep the height; just the difference in height,
i.e. the balance factor; this has to be modified on the path of
insertion even if you don’t perform rotations
Once you have performed a rotation (single or double) you won’t
need to go back up the tree

31. 31
Single Rotation
RotateFromRight(n : reference node pointer) {
p : node pointer;
p := n.right;
n
n.right := p.left;
p.left := n;
n := p
}
You also need to
X
modify the heights
or balance factors
of n and p
Y Z
Insert

32. 32
Double Rotation
• Implement Double Rotation in two lines.
DoubleRotateFromRight(n : reference node pointer) {
????
n
}
X
V W
Z

33. 33
Insertion in AVL Trees
• Insert at the leaf (as for all BST)
› only nodes on the path from insertion point to
root node have possibly changed in height
› So after the Insert, go back up to the root
node by node, updating heights
› If a new balance factor is 2 or –2, adjust tree
by rotation around the node

34. 34
Insert in BST
Insert(T : reference tree pointer, x : element) :
integer {
if T = null then
T := new tree; T.data := x; return 1;//the
links to
//children are null
case
T.data = x : return 0; //Duplicate do nothing
T.data > x : return Insert(T.left, x);
T.data < x : return Insert(T.right, x);
endcase
}

35. 35
Insert in AVL trees
Insert(T : reference tree pointer, x : element) : {
if T = null then
{T := new tree; T.data := x; height := 0; return;}
case
T.data = x : return ; //Duplicate do nothing
T.data > x : Insert(T.left, x);
if ((height(T.left)- height(T.right)) = 2){
if (T.left.data > x ) then //outside case
T = RotatefromLeft (T);
else //inside case
T = DoubleRotatefromLeft (T);}
T.data < x : Insert(T.right, x);
code similar to the left case
Endcase
T.height := max(height(T.left),height(T.right)) +1;
return;
}

36. Example of Insertions in an AVL Tree
36
1
0
2
20
10 30
25
0
0
35
Insert 5, 40

37. Example of Insertions in an AVL Tree
0 1
37
1
0
2
20
10 30
25
1
0
35
0
5
20
10 30
25
1
5 35
40
0
0
2
3
Now Insert 45

38. Single rotation (outside case)
1
0 0
38
2
0
3
20
10 30
25
1
2
35
0
5
20
10 30
25
1
5 40
40
0
0
0
1
2
3
45
Imbalance
35 45
Now Insert 34

39. 0
1
39
Double rotation (inside case)
3
0
3
20
10 30
25
1
2
40
0
5
20
10 35
30
1
5 40
45
0 1
2
3
Imbalance
45
Insertion of 34
35
34
0
0
1 0 25 34

40. 40
AVL Tree Deletion
• Similar but more complex than insertion
› Rotations and double rotations needed to
rebalance
› Imbalance may propagate upward so that
many rotations may be needed.

41. Pros and Cons of AVL Trees
Arguments for AVL trees:
1. Search is O(log N) since AVL trees are always balanced.
2. Insertion and deletions are also O(logn)
3. The height balancing adds no more than a constant factor to the
41
speed of insertion.
Arguments against using AVL trees:
1. Difficult to program & debug; more space for balance factor.
2. Asymptotically faster but rebalancing costs time.
3. Most large searches are done in database systems on disk and use
other structures (e.g. B-trees).
4. May be OK to have O(N) for a single operation if total run time for
many consecutive operations is fast (e.g. Splay trees).

42. 42
Double Rotation Solution
DoubleRotateFromRight(n : reference node pointer) {
RotateFromLeft(n.right);
RotateFromRight(n);
}
X
n
V W
Z