1. Representation of an AVL Node
class AVLNode
{ < declarations for info stored in node, e.g. int info; >
AVLnode left;
AVLnode right;
int height;
int height(AVLNode T) {
return T == null? -1 : T.height; } … };
COSC 2P03 Week 5

2. AVL insert AVLNode insert(AVLNode T, AVLNode newNode) { if(T == null)
T = newNode;
else if(newNode.info < T.info) // insert in left subtree
T.left = insert(T.left, newNode);
if(height(T.left) - height(T.right) == 2)
if(newNode.info < T.left.info) //left subtree of T.left
T = rotateWithLeftChild(T);
else //right subtree of T.left
T = doubleWithLeftChild(T); }
else // insert in right subtree
T.right = insert(T.right, newNode);
if(height(T.right) - height(T.left) == 2)
if(newNode.info > T.right.info) // right subtree of T.right
T = rotateWithRightChild(T);
else // left subtree of T.right
T = doubleWithRightChild(T);
T.height = max(height(T.left), height(T.right)) + 1;
return T;
COSC 2P03 Week 5

3. Single rotation – left-left
Insertion in left subtree of k2.left caused an imbalance at k2: need to rebalance from k2 down.
AVLNode rotateWithLeftChild(AVLNode k2) {
AVLNode k1 = k2.left;
k2.left = k1.right;
k1.right = k2;
k2.height = max(height(k2.left,
height(k2.right)) + 1;
k1.height = max(height(k1.left),
k2.height) + 1;
return k1; }
Note: right-right case is symmetric to above (left↔right).
COSC 2P03 Week 5

4. Double rotation – right-left
Insertion in right subtree of left child caused imbalance at k3: need to rebalance from k3 down.
AVLNode doubleWithLeftChild(AVLNode k3) {
k3.left = rotateWithRightChild(k3.left);
return rotateWithLeftChild(k3); }
Note: left-right case is symmetric to above (left↔right).
COSC 2P03 Week 5

5. B Trees (section 4.7 of textbook)
Commonly used for organizing data on disk
Disk access time (time to read or write a block) dominates cost
Very important to minimize number of disk accesses
Some notes on terminology:
This textbook uses the name B tree, but elsewhere they are known as B+ trees
In this textbook, the order of a B tree is the maximum number of children per index node.
Elsewhere, order refers to the minimum number of children in index nodes other than the root.
COSC 2P03 Week 5

6. B tree definitions A B-tree of order M is an M-ary tree such that:
Data items are only in the leaves
Non-leaf (index) nodes store up to M-1 keys:
key i determines the smallest possible key in subtree i+1.
The root is either a leaf or has between 2 and M children
Every node other than the root is at least half-full:
All non-leaf nodes (except root) have at least M/2 children
All leaves are at the same depth and have between L/2 and L records.
COSC 2P03 Week 5

7. B tree and Binary Search Tree comparison
Binary search trees: nodes have 0, 1 or 2 children and 1 key
B-trees: non-leaf nodes have up to M children: a node with d keys has d+1 children
Binary search trees: data is stored in both leaf and non-leaf nodes, and for any given index node N:
N’s left subtree contains only items with keys < N’s key
N’s right subtree contains only items with keys > N’s key
B-trees: data is stored only in leaf nodes. Non-leaf nodes contain up to M-1 keys (k1, …, kM-1) and:
Subtree to left of k1 contains only items with keys <k1
Subtree between ki and ki+1 contains only items with keys <ki+1 and ≥ki
Subtree to right of kM-1 contains only items with keys ≥kM-1
COSC 2P03 Week 5

8. B-tree Example M=5 and L=3
100
150
200 10 30 50
110
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260 3 10 30 50
100
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260 7 15 35 80
105
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125
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190
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270 20 90
170
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275
COSC 2P03 Week 5

9. B-tree example: insert 40
100
150
200 10 30 50
110
120
130
160
180
220
240
260 3 10 30 50
100
110
120
130
150
160
180
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260 7 15 35 80
105
115
125
140
155
165
190
210
225
250
270 20 40 90
170
230
275
COSC 2P03 Week 5

10. B-tree example: insert 70
100
150
200 10 30 50 80
110
120
130
160
180
220
240
260 3 10 30 50 80
100
110
120
130
150
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180
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260 7 15 35 70 90
105
115
125
140
155
165
190
210
225
250
270 20 40
170
230
275
COSC 2P03 Week 5

11. B-tree example: insert 2530
100
150
200 10 20 50 80
110
120
130
160
180
220
240
260 3 10 20 30 50 80
100
110
120
130
150
160
180
200
220
240
260 7 15 25 35 70 90
105
115
125
140
155
165
190
210
225
250
270 40
170
230
275
COSC 2P03 Week 5

12. B-tree example: insert 23530
100
150
200 10 20 50 80
110
120
130
160
180
220
230
240
260 3 10 20 30 50 80
100
110
120
130
150
160
180
200
220
230
240
260 7 15 25 35 70 90
105
115
125
140
155
165
190
210
225
235
250
270 40
170
275
COSC 2P03 Week 5

13. B-tree example: insert 280
150 30
100
200
240 10 20 50 80
110
120
130
160
180
220
230
260
275 3 10 20 30 50 80
100
110
120
130
150
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180
200
220
230
240
260
275 7 15 25 35 70 90
105
115
125
140
155
165
190
210
225
235
250
270
280 40
170
COSC 2P03 Week 5

14. Heaps
A heap is a binary tree that satisfies all of the following properties:
Structure property: It is a complete binary tree
Heap-order property: Every node satisfies the heap condition:
The key of every node n must be smaller than (or equal to) the keys of its children, i.e.
n.info  n.left.info and
n.info  n.right.info
COSC 2P03 Week 5