1. Cinda Heeren / Geoffrey Tien
AVL Trees
Properties
Insertion
October 17, 2017
Cinda Heeren / Geoffrey Tien

2. Cinda Heeren / Geoffrey Tien
BST rotation g g e h b h b f i a e i a c c f d d
Rotations preserve the in-order BST property, while altering the tree structure
we can use rotations to bring imbalanced trees back into balance
October 12, 2017
Cinda Heeren / Geoffrey Tien

3. Cinda Heeren / Geoffrey Tien
AVL trees
An AVL tree is a balanced BST
Each node's left and right subtrees differ in height by at most 1
Rebalancing via rotations occurs when an insertion or removal causes excessive height difference
AVL tree nodes contain extra information to support this height information 43 19 63 57 60 50 78 38 4 21
October 17, 2017
Cinda Heeren / Geoffrey Tien

4. Cinda Heeren / Geoffrey Tien
AVL nodes
enum balance_type {LEFT_HEAVY = -1, BALANCED = 0, RIGHT_HEAVY = +1};
class AVLNode {
public:
int data; // or template type
AVLNode left;
AVLNode right;
balance_type balance;
AVLNode(int value) { ... }
AVLNode(int val, AVLNode left1, AVLNode right1) { ... } }
AVLNode is almost the same as a binary tree node
additional balance field indicates that state of subtree balance at that node
October 17, 2017
Cinda Heeren / Geoffrey Tien

5. Cinda Heeren / Geoffrey Tien
AVL imbalance
Balanced trees: 3 3 7 12 16 3 7 12 19 3 16 27 31 44
Imbalanced trees: 12 19 3 16 27 31 44 21 24 12 16 3 7 9 3 7 5
October 17, 2017
Cinda Heeren / Geoffrey Tien

6. Cinda Heeren / Geoffrey Tien
AVL imbalance
4 cases of imbalance C B A C A B A B C A C B
LL imbalance
LR imbalance
RR imbalance
RL imbalance
Solve with a right rotation around C
Left rotation around A, becomes LL case
Symmetric to left imbalance cases
October 17, 2017
Cinda Heeren / Geoffrey Tien

7. Cinda Heeren / Geoffrey Tien
AVL insertion
Maintaining balance
The best way to keep a tree balanced, is to never let it become imbalanced!
AVL insertion begins with ordinary BST insertion (i.e. a leaf node) followed by rotations to maintain balance
i.e. AVL properties are satisfied before and after insertion
if the balance attribute of a subtree's root node becomes critical (-2 or +2) as a result of inserting the new leaf, rebalance it!
October 17, 2017
Cinda Heeren / Geoffrey Tien

8. Cinda Heeren / Geoffrey Tien
AVL insertion
Pseudocode
if root is NULL
Create new node containing item, assign root to it, and return true
else if item is equal to root->data
item exists already, return false
else if item < root->data
Recursively insert the item into the left subtree
if height of left subtree has increased (increase variable is true)
balance--;
if balance == 0, reset increase variable to false
if balance < -1
reset increase variable to false
perform rebalanceLeft
else if item > root->data
(symmetric to left subtree case, incrementing balance)
rebalanceLeft and rebalanceRight are the rotations to correct the 4 imbalance cases
October 17, 2017
Cinda Heeren / Geoffrey Tien

9. Cinda Heeren / Geoffrey Tien
AVL insertion example
Insert(65) 47 32 71 65 93
October 17, 2017
Cinda Heeren / Geoffrey Tien

10. Cinda Heeren / Geoffrey Tien
AVL insertion example
Insert(65)
Insert(82) 47
RR imbalance 32 71 OK 65 93 OK 82 OK
October 17, 2017
Cinda Heeren / Geoffrey Tien

11. Cinda Heeren / Geoffrey Tien
AVL insertion example
Insert(65)
Insert(82) 71
Insert(87) 47 93
LR imbalance 32 65 82 OK 87 OK
October 17, 2017
Cinda Heeren / Geoffrey Tien

12. Cinda Heeren / Geoffrey Tien
AVL insertion example
Insert(65)
Insert(82) 71 OK
Insert(87) 47 87 OK 32 65 82 93
October 17, 2017
Cinda Heeren / Geoffrey Tien

13. Cinda Heeren / Geoffrey Tien
AVL tree height
The height of an AVL tree containing 𝑛 key values is 𝑂 log 𝑛
Intuition:
For a fixed height ℎ, a tree containing fewer nodes has a larger height-to-node ratio
𝑛=15
𝑛=4
ℎ=3= log 𝑛 =𝑂( log 𝑛 )
ℎ=3=𝑛−1=𝑂 𝑛
Attempt to achieve the worst ratio by making an AVL tree of height ℎ with the minimum number of nodes
October 17, 2017
Cinda Heeren / Geoffrey Tien

14. Cinda Heeren / Geoffrey Tien
AVL tree height
Theorem:
The height of an AVL tree with 𝑛 nodes is 𝑂 log 𝑛
Proof:
Let 𝑁 ℎ represent the minimum number of nodes in an AVL tree of height ℎ
Since the AVL property must be satisfied at every node, the children of such a tree must also be minimal, and the height difference between the children must be 1
Thus 𝑁 ℎ =1+ 𝑁 ℎ−1 + 𝑁 ℎ−2 1
Nh-2
Nh-1
October 17, 2017
Cinda Heeren / Geoffrey Tien

15. Cinda Heeren / Geoffrey Tien
AVL tree height
𝑁 ℎ =1+ 𝑁 ℎ−1 + 𝑁 ℎ−2
𝑁 ℎ−1 =1+ 𝑁 ℎ−2 + 𝑁 ℎ−3
𝑁 ℎ =1+ 1+ 𝑁 ℎ−2 + 𝑁 ℎ−3 + 𝑁 ℎ−2
=2+2 𝑁 ℎ−2 + 𝑁 ℎ−3 >2 𝑁 ℎ−2
𝑁 ℎ−2 =1+ 𝑁 ℎ−3 + 𝑁 ℎ−4
=2+2 𝑁 ℎ−4 + 𝑁 ℎ−5 >2 𝑁 ℎ−4
𝑁 ℎ >2∙2 𝑁 ℎ−4
𝑁 ℎ >2∙2∙2 𝑁 ℎ−6
How many times can we subtract 2 from ℎ before we reach 0?
ℎ 2 times.
𝑁 ℎ >2∙2∙2∙2 𝑁 ℎ−8 …
𝑁 ℎ > 2 ℎ 2
ℎ< log 𝑁 ℎ
ℎ∈𝑂 log 𝑛
October 17, 2017
Cinda Heeren / Geoffrey Tien

16. Cinda Heeren / Geoffrey Tien
Exercise
Draw the AVL tree resulting from the following sequence of insertions: 71, 13, 65, 22, 49, 37, 45
Note the height recorded/updated at each node
What would an ordinary BST look like with the same sequence of insertions?
October 17, 2017
Cinda Heeren / Geoffrey Tien

17. Readings for this lesson
Koffman
Chapter 11.2 (AVL trees)
AVL visualisations:
(AVL tree)
Next class: AVL removal
October 17, 2017
Cinda Heeren / Geoffrey Tien