2. AVL Trees
Balance
Balance == height(left subtree) - height(right subtree)
zero everywhere perfectly balanced
small everywhere balanced enough
Balance between -1 and 1 everywhere
maximum height of ~1.44 log n
t
5
7
3. AVL Trees
AVL Tree
Dictionary Data Structure
4
12
10
6
2
11
5
8
14
13
7 9
Binary search tree properties
binary tree property
search tree property
Balance property
balance of every node is:
-1 b 1
result:
depth is (log n)
15
4. AVL Trees
An AVL Tree
15
9
2 12
5
10
20
17
0
0
1
0
0
1 2
3 10
3
data
height
children
30
0
6. AVL Trees
Good Insert Case:
Balance Preserved
Good case: insert middle, then small,then tall
Insert(middle)
Insert(small)
Insert(tall)
M
S T
0
0
1
7. AVL Trees
Bad Insert Case #1:
Left-Left or Right-Right Imbalance
Insert(small)
Insert(middle)
Insert(tall)
T
M
S
0
1
2
BC#1 Imbalance caused by either:
Insert into left child’s left subtree
Insert into right child’s right subtree
9. AVL Trees
General Bad Case #1
a
X
Y
b
Z
h h - 1
h + 1 h - 1
h + 2
a
X
Y
b
Z
h-1 h - 1
h h - 1
h + 1
Note: imbalance is left-left
10. AVL Trees
Single Rotation
Fixes Case #1 Imbalance
Height of left subtree same as it was before insert!
Height of all ancestors unchanged
We can stop here!
a
X
Y
b
Z
a
X
Y
b
Z
h h - 1
h + 1 h - 1
h + 2
h
h - 1
h
h - 1
h + 1
11. AVL Trees
Bad Insert Case #2:
Left-Right or Right-Left Imbalance
Insert(small)
Insert(tall)
Insert(middle)
M
T
S
0
1
2
Will a single rotation fix this?
BC#2 Imbalance caused by either:
Insert into left child’s right subtree
Insert into right child’s left subtree
13. AVL Trees
General Bad Case #2
a
X
b
Z
h
h + 1 h - 1
h + 2
a
X
b
Z
h-1 h - 1
h h - 1
h + 1
Note: imbalance is left-right
h-1
Y
Y
14. AVL Trees
Double Rotation
Fixes Case #2 Imbalance
Initially: insert into either X or Y unbalances tree (root balance goes to 2 or -2)
“Zig zag” to pull up c – restores root height to h+1, left subtree height to h
a
Z
b
W
c
X Y
a
Z
b
W
c
X Y
h
h - 1?
h - 1
h - 1
h + 2
h + 1
h - 1
h - 1
h
h + 1
h
h - 1?
15. AVL Trees
AVL Insert Algorithm
Find spot for value
Hang new node
Search back up looking for imbalance
If there is an imbalance:
case #1: Perform single rotation
case #2: Perform double rotation
Done!
(There can only be one imbalance!)
23. AVL Trees
AVL Insert Algorithm Revisited
Recursive
1. Search downward for
spot
2. Insert node
3. Unwind stack,
correcting heights
a. If imbalance #1,
single rotate
b. If imbalance #2,
double rotate
Iterative
1. Search downward for
spot, stacking
parent nodes
2. Insert node
3. Unwind stack,
correcting heights
a. If imbalance #1,
single rotate and
exit
b. If imbalance #2,
double rotate and
exit
24. AVL Trees
Single Rotation Code
void RotateRight(Node *& root) {
Node * temp = root->right;
root->right = temp->left;
temp->left = root;
root->height =
max(root->right->height,
root->left->height) + 1;
temp->height =
max(temp->right->height,
temp->left->height) + 1;
root = temp;
}
X
Y
Z
root
temp
25. AVL Trees
Double Rotation Code
void DoubleRotateRight(Node *& root) {
RotateLeft(root->right);
RotateRight(root);
}
a
Z
b
W
c
X
Y
a
Z
c
b
X
Y
W
First Rotation
34. AVL Trees
Deletion with Propagation
We get to choose whether
to single or double rotate!
20
5
2
17
3
10
30
12
0 2
2
0
1 3
4
33
15
13
1
0 0
1
11
0
18
0
What different about this case?
37. AVL Trees
AVL Deletion Algorithm
Recursive
1. Search downward for
node
2. Delete node
3. Unwind, correcting
heights as we go
a. If imbalance #1,
single rotate
b. If imbalance #2
(or don’t care),
double rotate
Iterative
1. Search downward for
node, stacking
parent nodes
2. Delete node
3. Unwind stack,
correcting heights
a. If imbalance #1,
single rotate
b. If imbalance #2
(or don’t care)
double rotate
38. AVL Trees
Building an AVL Tree
Input: sequence of n keys (unordered)
19 3 4 18 7
Insert each into initially empty AVL tree
But, suppose input is already sorted …
3 4 7 18 19
Can we do better than O(n log n)?
1 1
( lo
l g )
og log
n n
i i
O n n
i n
39. AVL Trees
AVL BuildTree
8 10 15 20 30 35 40
5
17
17
8 1015
5 20303540
Divide & Conquer
Divide the problem into
parts
Solve each part recursively
Merge the parts into a
general solution
How long does
divide & conquer take?
41. AVL Trees
BuildTree Analysis
(Approximate)
T(1) = 1
T(n) = 2T(n/2) + 1
T(n) = 2(2T(n/4)+1) + 1
T(n) = 4T(n/4) + 2 + 1
T(n) = 4(2T(n/8)+1) + 2 + 1
T(n) = 8T(n/8) + 4 + 2 + 1
T(n) = 2kT(n/2k) +
let 2k = n, log n = k
T(n) = nT(1) +
T(n) = (n)
k
i
i
1
2
2
1
n
i
i
log
1
2
2
1
42. AVL Trees
Precise Analysis: T(0) = b
T(n) = T( ) + T( ) + c
By induction on n:
T(n) = (b+c)n + b
Base case:
T(0) = b = (b+c)0 + b
Induction step:
T(n) = (b+c) + b +
(b+c) + b + c
= (b+c)n + b
QED: T(n) = (b+c)n + b = (n)
2
1
n
2
1
n
2
1
n
2
1
n
BuildTree Analysis
(Exact)
1
2
1
2
1
n
n
n
43. AVL Trees
Thinking About AVL
Observations
+ Worst case height of an AVL tree is about 1.44 log n
+ Insert, Find, Delete in worst case O(log n)
+ Only one (single or double) rotation needed on
insertion
+ Compatible with lazy deletion
- O(log n) rotations needed on deletion
- Height fields must be maintained (or 2-bit balance)
Coding complexity?
44. AVL Trees
Alternatives to AVL Trees
Weight balanced trees
keep about the same number of nodes in each subtree
not nearly as nice
Splay trees
“blind” adjusting version of AVL trees
no height information maintained!
insert/find always rotates node to the root!
worst case time is O(n)
amortized time for all operations is O(log n)
mysterious, but often faster than AVL trees in practice
(better low-order terms)
