1. AVL Trees binary tree for every node x, define its balance factor
balance factor of x = height of left subtree of x
– height of right subtree of x
balance factor of every node x is – 1, 0, or 1
log2 (n+1) <= height <= 1.44 log2 (n+2)

2. Example AVL Tree 1 -1 10 7 8 3 5 30 40 20 25 35 45 60

3. put(9) -1 10 1 1 7 40 -1 -1 1 45 3 8 30
Some balance factors on insert path change. The height of a subtree on this insert path can change by at most 1. So, bf of a node on the insert path can change by at most 1.
When a bf changes from 0 to +1 or –1 the subtree height increases and we need to go further up the tree adjusting balance factors. -1 60 35 1 9 20 5 25

4. put(29) -1 10 1 1 7 40 -1 1 45 3 8 30 -1 -2 60 35 1 20 5
White node is the A node, nearest ancestor of newly inserted node whose bf becomes +2 or –2.
RR imbalance => new node is in right subtree of right subtree of white node (node with bf = –2) -1 25 29

5. put(29) -1 10 1 1 7 40 -1 1 45 3 8 30 60 35 1 25 5 20 29
RR rotation.

6. Insert/Put
Following insert/put, retrace path towards root and adjust balance factors as needed.
Stop when you reach a node whose balance factor becomes 0, 2, or –2, or when you reach the root.
The new tree is not an AVL tree only if you reach a node whose balance factor is either 2 or –2.
In this case, we say the tree has become unbalanced.

7. A-Node
Let A be the nearest ancestor of the newly inserted node whose balance factor becomes +2 or –2 following the insert.
Balance factor of nodes between new node and A is 0 before insertion.

8. Balance Factors Between New Node and A Are 01 -1 10 7 8 3 5 30 40 20 25 35 45 60
Insert 6

9. Imbalance Types
RR … newly inserted node is in the right subtree of the right subtree of A.
LL … left subtree of left subtree of A.
RL… left subtree of right subtree of A.
LR… right subtree of left subtree of A.

10. LL Rotation Subtree height is unchanged.
Before insertion. 1 A B BL BR AR h A B
B’L BR AR
After insertion.
h+1 h B 2
B’L
h+1 A 1 BR h AR h
After rotation.
Subtree height is unchanged.
No further adjustments to be done.

11. LR Rotation (case 1) Subtree height is unchanged.
Before insertion. 1 A B A B
After insertion. C C 2 B A -1
After rotation.
B is a leaf prior to the insert.
Subtree height is unchanged.
No further adjustments to be done.

12. LR Rotation (case 2) Subtree height is unchanged.BL CR AR h
h-1
C’L C 1 A B BL CR AR h
h-1 CL C C 2 B A -1 -1 BL h
C’L h CR
h-1 AR h 1
B is not a leaf prior to the insert, the insert takes place in the left subtree of C.
Subtree height is unchanged.
No further adjustments to be done.

13. LR Rotation (case 3) Subtree height is unchanged.BL
C’R AR h CL
h-1 C -1 2 1 A B BL CR AR h
h-1 CL C C A
C’R h AR B BL CL
h-1 1
Subtree height is unchanged.
No further adjustments to be done.

14. Single & Double Rotations
LL and RR
Double
LR and RL
LR is RR followed by LL
RL is LL followed by RR

15. LR Is RR + LL A C CL C’R AR h BL B After RR rotation. h-1 A B BL C’R2
After RR rotation.
h-1 A B BL
C’R AR h CL
h-1 C -1 2
After insertion. C A
C’R h AR B BL CL
h-1 1
After LL rotation.

16. Remove An Element Remove 8. 1 -1 10 7 8 3 5 30 40 20 25 35 45 601 -1 10 7 8 3 5 30 40 20 25 35 45 60
Remove 5; no traceback needed.
Remove 8.

17. Remove An Element q Let q be parent of deleted node.2 -1 1 10 7 3 5 30 40 20 25 35 45 60 q
No q => no rebalancing to be done.
Let q be parent of deleted node.
Retrace path from q towards root.

18. New Balance Factor Of q Deletion from left subtree of q => bf--.
Deletion from right subtree of q => bf++.
New balance factor = 1 or –1 => no change in height of subtree rooted at q.
New balance factor = 0 => height of subtree rooted at q has decreased by 1.
New balance factor = 2 or –2 => tree is unbalanced at q.
Initially q is the parent of the deleted node. At this time, one subtree of q is empty and the other is not empty. As q moves up towards the root, neither subtree of q is empty.

19. Imbalance Classification
Let A be an ancestor (q) of the deleted node whose balance factor has become 2 or –2 following a deletion.
Deletion from left subtree of A => type L.
Deletion from right subtree of A => type R.
Type R => new bf(A) = 2.
So, old bf(A) = 1.
So, A has a left child B.
bf(B) = 0 => R0.
bf(B) = 1 => R1.
bf(B) = –1 => R-1.

20. R0 Rotation Subtree height is unchanged.
Before deletion. 1 A B BL BR AR h A B BL BR
A’R
After deletion. h
h-1 2 B A
After rotation. BR h
A’R
h-1 BL 1 -1
Subtree height is unchanged.
No further adjustments to be done.
Similar to LL rotation.

21. R1 Rotation Subtree height is reduced by 1.
Before deletion. 1 A B BL BR AR h
h-1 A B BL BR
A’R
After deletion. h
h-1 1 2 B A
After rotation. BR
h-1
A’R BL h
Subtree height is reduced by 1.
Must continue on path to root.
Similar to LL and R0 rotations.

22. R-1 Rotation New balance factor of A and B depends on b.BL CR
A’R
h-1 CL C b -1 2 1 -1 A B BL CR AR
h-1 h b CL C C A CR
A’R
h-1 B BL CL
New balance factor of A and B depends on b.
Subtree height is reduced by 1.
Must continue on path to root.
Similar to LR.

23. Number Of Rebalancing Rotations
At most 1 for an insert.
O(log n) for a delete.

24. Rotation Frequency Insert random numbers.
No rotation … 53.4% (approx).
LL/RR … 23.3% (approx).
LR/RL … 23.2% (approx).
Frequencies do not sum to 100% because of rounding errors in reporting frequencies to one decimal place.