Trees

About This Presentation

Title:

Trees

Description:

You've probably seen family trees like the one at the right. 2. Building a Tree ... the ancestors (parents, grand parents and so on) of one person in a family tree ... – PowerPoint PPT presentation

Number of Views:100

Avg rating:3.0/5.0

Slides: 158

Provided by: lowel3

Category:

more less

Transcript and Presenter's Notes

Title: Trees

1
Trees

A tree is a data structure similar to a linked
list, except there is more than one pointer
You've probably seen family trees like the one at
the right

2
Building a Tree

To model this in Java we would need a class like

class TreeNode String myName treeNode
pMom, pDad treeNode (String sN, treeNode pL,
treeNode pR) myName sN pMom pL pDad
pR
3
Building a Tree

Since we don't know who Ellen and John's parents
are, we'll use null

TreeNode pEllen new TreeNode("Ellen
Rimbauer",null,null) TreeNode pJohn new
TreeNode("John Rimbauer",null,null)
4
Building a Tree

Adam's parents are Ellen and John

TreeNode pEllen new TreeNode("Ellen
Rimbauer",null,null) TreeNode pJohn new
TreeNode("John Rimbauer",null,null) TreeNode
pAdam new TreeNode("Adam
Rimbauer",pEllen,pJohn)
5
Building a Tree

April's parents are also Ellen and John
Steven's dad is Adam, but we don't know his
mother

TreeNode pAdam new TreeNode("Adam
Rimbauer",pEllen,pJohn) TreeNode pApril
new TreeNode("April Rimbauer",pEllen,pJohn) TreeN
ode pSteven new TreeNode("Steven
Rimbauer",null, pAdam)
6
An better encapsulated tree class

public class TreeNode
private Object value //note Object
private TreeNode left
private TreeNode right
public TreeNode(Object initValue, TreeNode
initLeft, TreeNode initRight)
value initValue left initLeft
right initRight
public Object getValue() return value
public TreeNode getLeft() return left
public TreeNode getRight() return right
public void setValue(Object theNewValue)
value theNewValue
public void setLeft(TreeNode theNewLeft)
left theNewLeft
public void setRight(TreeNode theNewRight)
right theNewRight

7
Methods that operate on Trees

Just like a linked list, we can make a
fill-in-the-blank outline for methods that
operate on treeNodes
. . . funForTreeNode (TreeNode root)
if (root null)
. . . . .
else
. . . . .root.getValue(). . . .
. . . FunForTree(root.getLeft()) . . . .
. . . FunForTree(root.getRight()). . . .

8
Methods that operate on Trees

We have three possible arrangements for the
recursive calls Pre-Order
. . . funForTreeNode (TreeNode root)
if (root null)
. . . . .
else
. . . . .root.getValue(). . . .
. . . . FunForTree(root.getLeft()). . . .
. . . . FunForTree(root.getRight()) . . .

9
Methods that operate on Trees

We have three possible arrangements for the
recursive calls In-Order
. . . funForTreeNode (TreeNode root)
if (root null)
. . . . .
else
. . . . FunForTree(root.getLeft()). . . .
. . . . .rootgetValue(). . . .
. . . . FunForTree(root.getRight()) . . .

10
Methods that operate on Trees

We have three possible arrangements for the
recursive calls Post-Order
. . . funForTreeNode (TreeNode root)
if (root null)
. . . . .
else
. . . . FunForTree(root.getLeft()). . . .
. . . . FunForTree(root.getRight()) . . .
. . . . .root.getValue(). . . .

11
Problem Write a method that counts all the
ancestors (parents, grand parents and so on) of
one person in a family tree

int CountParents (treeNode root)
if (root null)
. . . . .
else
. . . . .root.getValue(). . . .
. . . . CountParents(root.getLeft()). . .
. . . . CountParents(root.getRight()) .
. .

12
Problem Write a method that counts all the
ancestors (parents, grand parents and so on) of
one person in a family tree

int CountParents (TreeNode root)
if (root null)
return 0
else
. . . . root.getValue()
. . . . CountParents(root.getLeft()) . .
. .
. . . . CountParents(root.getRight()). .
. .

13
Problem Write a method that counts all the
ancestors (parents, grand parents and so on) of
one person in a family tree

int CountParents (TreeNode root)
if (root null)
return 0
else
return 1
CountParents(root.getLeft())
CountParents(root.getRight())

14
Problem Write a method that counts all the
ancestors (parents, grand parents and so on) of
one person in a family tree

int CountParents (TreeNode root)
if (root null)
return 0
else
return 1
CountParents(root.getLeft())
CountParents(root.getRight())
The problem with this solution is that the person
counts as their own ancestor

15
Problem Write a method that counts all the
ancestors (parents, grand parents and so on) of
one person in a family tree

We can write a helper method to fix this problem
int CountAncestors(TreeNode root)
return CountParents(root)-1
//or
int CountAncestors(TreeNode root)
return CountParents(root.pMom)
CountParents(root.pDad)

16
Binary Search Trees

Definition A BST is a tree where each node has
two pointers and for any node, the element in the
node is larger than all elements in this node's
left subtree and smaller than all elements in
this node's right subtree
Huh?

17
Binary Search Trees

Some of these are BSTs, some aren't

10
10
10
5
5
15
5
11
4
11
4
11
4
7
16
12
4
7
12
7
12
3
7
5
10
5
9
4
11
5
11
3
7
4
7
12
9
18
Binary Search Trees

These are BSTs

10
10
5
10
5
15
5
11
4
11
4
11
4
7
16
12
4
7
12
3
7
7
12
5
9
19
Binary Search Trees

These aren't BSTs

5
10
4
11
5
11
3
7
4
7
12
9
20
Building a BST

Let's say you want to build a BST from the
numbers 2 1 8 7 3

21
Building a BST

Let's say you want to build a BST from the
numbers 2 1 8 7 3

2
22
Building a BST

Let's say you want to build a BST from the
numbers 2 1 8 7 3

2
1
23
Building a BST

Let's say you want to build a BST from the
numbers 2 1 8 7 3

2
1
8
24
Building a BST

Let's say you want to build a BST from the
numbers 2 1 8 7 3

2
1
8
7
25
Building a BST

Let's say you want to build a BST from the
numbers 2 1 8 7 3

2
1
8
7
3
26
Tree Vocabulary
E
B
H
A
C
I
G
D

BST are usually displayed "upside down", the root
is at the top
Each element is a node
Any node whose left and right subtrees are empty
is a leaf
The level of a node counts down from the root A,
C, G and I are all at level 2, the level of the
tree is 3
The height of a tree is the number of nodes on
the longest path from root to leaf

27
Tree Vocabulary
E
B
H
A
C
I
G
D

A balanced tree has approximately the same number
of nodes in the left and right subtrees.
A full binary tree has every leaf on the same
level, every nonleaf node is the parent of two
children
A complete binary tree is either full or full
through the next to last level with the leaves as
far left as possible

28
Searching a BST
E
B
H
A
C
I
G
D

If a BST is reasonably balanced, finding a node
is quick and easy.
Let's say we are looking for 'D'.
Start at the root. If the node you are looking
for is smaller go left, otherwise, go right
Even though there are 8 nodes, the most nodes
we'll visit is 4--the height of the tree

29
Searching a BST

If a linked list has 1000 nodes, and you are
searching for a particular value, what is the
most nodes you would have to visit?

30
Searching a BST

If a linked list has 1000 nodes, and you are
searching for a particular value, what is the
most nodes you would have to visit?
1000

31
Searching a BST

If a linked list has 1000 nodes, and you are
searching for a particular value, what is the
most nodes you would have to visit?
1000
If a balanced BST has 1000 nodes, and you are
searching for particular value, what is the most
nodes you will have to visit?

32
Searching a BST

If a linked list has 1000 nodes, and you are
searching for a particular value, what is the
most nodes you would have to visit?
1000
If a balanced BST has 1000 nodes, and you are
searching for particular value, what is the most
nodes you will have to visit?
10

33
Big O notation

If a linked list has n nodes, at most we would
have to visit n nodes
O(n)
If a balanced BST has n nodes, at most we would
have to visit log2 n nodes
O(log n)

34
Traversing a tree

Pre-Order
Visit
Left
Right
In-Order
Left
Visit
Right
Post-Order
Left
Right
Visit

E
B
H
A
C
I
G
D
35
Traversing a tree

Pre-Order
Visit
Left
Right

E
B
H
A
C
I
G
D
36
Traversing a tree

Pre-Order
Visit
Left
Right
E B A C D H G I

E
B
H
A
C
I
G
D
37
Traversing a tree

In-Order
Left
Visit
Right

E
B
H
A
C
I
G
D
38
Traversing a tree

In-Order
Left
Visit
Right
A B C D E G H I

E
B
H
A
C
I
G
D
39
Traversing a tree

Post-Order
Left
Right
Visit

E
B
H
A
C
I
G
D
40
Traversing a tree

Post-Order
Left
Right
Visit
A D C B G I H E

E
B
H
A
C
I
G
D
41
Practice Quiz Question What will the three
traversals give for this tree?
5
4
11
3
7
42
A BST class

Typically we build a BST from two classes
TreeNode represents a single node and BSTree
represents the entire tree
class TreeNode
private Object value //note Object!
private TreeNode left
private TreeNode right
public TreeNode(Object initValue, TreeNode
initLeft, TreeNode initRight)
value initValue left initLeft
right initRight
public Object getValue() return value
public TreeNode getLeft() return left
public TreeNode getRight() return right
public void setValue(Object theNewValue)
value theNewValue
public void setLeft(TreeNode theNewLeft)
left theNewLeft
public void setRight(TreeNode theNewRight)

43
A BST class

The BSTree class has a variable for the root and
methods to count, display, find and delete the
nodes
Typically, for each operation is there a simple
method that "starts" the process, and a recursive
"helper" that does the work
class BSTree
private TreeNode root
public BSTree()
public int startCount()
private int count(TreeNode root)
//and lots more

44
A BST class

class BSTree
private TreeNode root
public BSTree()
public int startCount()
return count(root)
private int count(TreeNode root)
//and lots more

45
A BST class

class BSTree
private TreeNode root
public BSTree()
public int startCount()
return count(root)
private int count(TreeNode root)
if(root null)
. . .
else
. . .count(root.getLeft()). . .
. . .root.getValue(). . .
. . .count(root.getRight()). . .

46
A BST class

class BSTree
private TreeNode root
public BSTree()
public int startCount()
return count(root)
private int count(TreeNode root)
if(root null)
return 0
else
. . .count(root.getLeft()). . .
. . .root.getValue(). . .
. . .count(root.getRight()). . .

47
A BST class

class BSTree
private TreeNode root
public BSTree()
public int startCount()
return count(root)
private int count(TreeNode root)
if(root null)
return 0
else
return count(root.getLeft())
1
count(root.getRight())

48
A BST class

class BSTree
private TreeNode root
public BSTree()
public boolean startFind(Comparable data)
return find(root, data)
public boolean find(TreeNode root,
Comparable data)

49
A BST class

class BSTree
private TreeNode root
public BSTree()
public boolean startFind(Comparable data)
return find(root, data)
public boolean find(TreeNode root,
Comparable data)
if(root null)
. . .
else
. . .find(root.getLeft(),data). . .
. . .root.getValue(). . .
. . .find(root.getRight(),data). . .

50
A BST class

class BSTree
private TreeNode root
public BSTree()
public boolean startFind(Comparable data)
return find(root, data)
public boolean find(TreeNode root,
Comparable data)
if(root null)
return false
else
. . .find(root.getLeft(),data). . .
. . .root.getValue(). . .
. . .find(root.getRight(),data). . .

51
A BST class

class BSTree
private TreeNode root
public BSTree()
public boolean startFind(Comparable data)
return find(root, data)
public boolean find(TreeNode root,
Comparable data)
if(root null)
return false
else
return find(root.getLeft(),data)
data.compareTo(root.getValue()) 0
find(root.getRight(),data)

52
A BST class

class BSTree
private TreeNode root
public BSTree()
public boolean startFind(Comparable data)
return find(root, data)
public boolean find(TreeNode root,
Comparable data)
if(root null)
return false
else
if(data.compareTo(root.getValue())
return find(root.getLeft(),data)
. . .root.getValue(). . .

53
A BST class

class BSTree
private TreeNode root
public BSTree()
public boolean startFind(Comparable data)
return find(root, data)
public boolean find(TreeNode root,
Comparable data)
if(root null)
return false
else
if(data.compareTo(root.getValue())
return find(root.getLeft(),data)
if(data.compareTo(root.getValue() 0)

54
A BST class

class BSTree
private TreeNode root
public BSTree()
public boolean startFind(Comparable data)
return find(root, data)
public boolean find(TreeNode root,
Comparable data)
if(root null)
return false
else
if(data.compareTo(root.getValue())
return find(root.getLeft(),data)
if(data.compareTo(root.getValue() 0)

55
A BST class

class BSTree
private TreeNode root
public BSTree()
public boolean startFind(Comparable data)
return find(root, data)
public boolean find(TreeNode root,
Comparable data)
if(root null)
return false
else
if(data.compareTo(root.getValue())
return find(root.getLeft(),data)
else if(data.compareTo(

56
Problem add a trim method to the BSTree class.
trim removes all leaves in the tree
Before
After
56
56
-2
78
-2
4
4
6
1
6
16
16
7
57
Problem add a trim method to the BSTree class.
trim removes all leaves in the tree

public void startTrim()
trim(root)
private void trim(TreeNode root)
if(root null)
. . . .
else
. . . root.getValue(). . .
. . .trim(root.getLeft()). . .
. . .trim(root.getRight()) . . .

58
Problem add a trim method to the BSTree class.
trim removes all leaves in the tree

public void startTrim()
trim(root)
private void trim(TreeNode root)
if(root null)
return
else
. . . root.getValue(). . .
. . .trim(root.getLeft()). . .
. . .trim(root.getRight()) . . .

59
Problem add a trim method to the BSTree class.
trim removes all leaves in the tree

public void startTrim()
trim(root)
private void trim(TreeNode root)
if(root null)
return
else
//if left is leaf delete it
//if right is leaf delete it
trim(root.getLeft())
trim(root.getRight())

60
Problem add a trim method to the BSTree class.
trim removes all leaves in the tree

public void startTrim()
trim(root)
private void trim(TreeNode root)
if(root null)
return
else
//if left is leaf delete it
if(root.getLeft()!null
root.getLeft().getRight()null
root.getLeft().getLeft()null)
root.setLeft(null)
//if right is leaf delete it
trim(root.getLeft())
trim(root.getRight())

61
Problem add a trim method to the BSTree class.
trim removes all leaves in the tree

public void startTrim()
trim(root)
private void trim(TreeNode root)
if(root null)
return
else
//if left is leaf delete it
if(root.getLeft()!null
root.getLeft().getRight()null
root.getLeft().getLeft()null)
root.setLeft(null)
//if right is leaf delete it
if(root.getRight()!null
root.getRight().getRight()null

62
Problem add a trim method to the BSTree class.
trim removes all leaves in the tree

public void startTrim()
trim(root)
private void trim(TreeNode root)
if(root null)
return
else
//THIS DOESN'T WORK!!!
if(root.getLeft()null
root.getRight()null)
root null
trim(root.getLeft())
trim(root.getRight())

63
Deleting a node from a tree

Two steps
Locate Node to delete
Eliminate the Node
After locating the node, there are four
possibilities
If it is a leaf, make the parent node point to
null.
If it has one child on the right, make the parent
node point to the right child
If it has one child on the left, make the parent
node point to the left child.
If it has two children, the problem becomes much
harder to solve.

64
Deleting a node from a tree

Three methods
public void startDelete(Comparable target)
private TreeNode delete(TreeNode node, Comparable
target)
private TreeNode deleteTargetNode(TreeNode
target)
startDelete() calls delete()
delete() recursively finds the node and calls
deleteTargetNode()
deleteTargetNode() eliminates the node
code is in ICT Java curriculum lesson 36

65
What does this have to do with computer
programming?
66
Stacks

Think of a stack of plates at a buffet
The first plate that is placed on the stack is
the last plate to be taken off
A stack is called a LIFO, last in first out
structure (or sometimes FILO)
Placing something on the stack is called pushing
Taking it off is popping

67
Stacks

A PEZ dispenser is another example of a stack

68
The Stack interface

public interface Stack
boolean isEmpty()
void push(Object x)
Object pop()
Object peekTop()
To Use the ArrayStack class that implements this
interface, you will need to download ap.jar

69
What is the output?

import ap.
public class StackDemo
public static void main(String args)
Stack s new ArrayStack()
s.push(new Integer(5))
s.push(new Integer(3))
s.push(new Integer(2))
s.pop()
System.out.println(s.peekTop())
s.pop()
int nNum ((Integer)s.pop()).intValue()
System.out.println(nNum)

70
Problem write a program that uses a stack to
reverse a string

String sForward
"A Man! A Plan! A Canal! Panama!"
String sBackward new String("")
Stack s new ArrayStack()
for(int nI 0 nI
s.push(new Character(sForward.charAt(nI)))
while(!s.isEmpty())
sBackward s.pop()
System.out.println(sBackward)
//Displays !amanaP !lanaC A !nalP A !naM A

71
What is the output?

Stack s new ArrayStack()
int nX 2, nY 3
s.push(new Integer(nX))
s.push(new Integer(2nY))
s.push(new Integer(4))
nY ((Integer)s.pop()).intValue()
s.push(new Integer(7))
s.pop()
nX ((Integer)s.peekTop()).intValue()
while(!s.isEmpty())
System.out.println(s.pop())
System.out.println("nX is " nX)

72
The Queue interface

Think of the line of people at a buffet
The first person in line is the first person to
get their food
A queue is called a FIFO, first in first out
structure
To place something at the end of the queue is
called enqueue
Removing the element at the front of the queue is
called dequeue

73
The Queue interface

public interface Queue
boolean isEmpty()
void enqueue(Object x)
Object dequeue()
Object peekFront()

74
The Queue interface

import ap.
public class QueueDemo
public static void main(String args)
Queue q new ListQueue()
q.enqueue(new Integer(5))
q.enqueue(new Integer(3))
q.enqueue(new Integer(2))
q.dequeue()
System.out.println(q.dequeue())

75
What is the output of this program?

Queue q new ListQueue()
int nX 2, nY 3
q.enqueue(new Integer(nX))
q.enqueue(new Integer(2nY))
q.enqueue(new Integer(4))
nY ((Integer)q.dequeue()).intValue()
q.enqueue(new Integer(7))
q.dequeue()
nX ((Integer)q.peekFront()).intValue()
while(!q.isEmpty())
System.out.println(q.dequeue())
System.out.println("nX is " nX)

76
A Hospital Emergency Room

You've seen how stacks and queues model the way a
buffet works
People come to a hospital for treatment Should
they be stored in a queue (FIFO), or a stack
(LIFO), or something else? How do you determine
who gets treated first?

77
Priority Queues

A Priority Queue is a data structure (like a
linked list, stack, tree, queue, array, etc) for
storing a collection of items
Each node has data and a priority
A Priority Queue is NOT A QUEUE, but a complete
binary tree
Priority Queues are also called "Heaps"

78
heaps (priority queues)

There are two versions
min heaps (the value of every node is less than
or equal to the value in each of its children)
max heaps (the value is each of its children)
The AP exam uses a min heap
If you used a heap to store the names of patients
awaiting treatment, the person on the top of the
heap would be the person most in need

1
6
4
8
7
9
5
12
20
79
Priority Queues

One way to construct a priority queue is to store
each node in an array
The root is at index one (index zero is unused)
If a node at index k has children, the left child
is at 2 k and the right child is at 2 k 1
For example, the node with value 4 is at index 3
It's left child is at index 6 (23)
It's right child is at index 7 (231)

1
6
4
8
7
9
5
12
20
0 1 2 3 4 5 6 7 8 9
1 6 4 8 7 9 5 12 20
80
Priority Queues

A well designed heap allows
Rapid insertion of elements that arrive in
arbitrary order
Rapid retrieval of the item with the highest
priority
Insertion and deletion of elements in a heap are
O(log n)

81
Fixing the heap (from the bottom)

If we add or remove and element, we will have to
rearrange the nodes to keep it a heap (called
reheaping)
For example, let's add a new node with the value
2 to the heap

1
6
4
8
7
9
5
12
20
82
Fixing the heap (from the bottom)

The heap is no longer a min heap
It needs to be reheaped from the bottom up

1
6
4
8
7
9
5
12
20
2
83
Fixing the heap (from the bottom)

We'll start at the last node with children (which
is the number of nodes divided by 2)
We'll compare each node with it's children
If an element has a lower priority than one of
its children, we swap it with the child with the
highest priority

1
6
4
8
7
9
5
12
20
2
84
Fixing the heap (from the bottom)

We'll start at the last node with children (which
is the number of nodes divided by 2)
We'll compare each node with it's children
If an element has a lower priority than one of
its children, we swap it with the child with the
highest priority

1
6
4
8
2
9
5
12
20
7
85
Fixing the heap (from the bottom)

We'll start at the last node with children (which
is the number of nodes divided by 2)
We'll compare each node with it's children
If an element has a lower priority than one of
its children, we swap it with the child with the
highest priority

1
2
4
8
6
9
5
12
20
7
86
Fixing the heap (from the top)

If we remove an element, we'll move each element
down one position in the array

1
2
4
8
6
9
5
12
20
7
0 1 2 3 4 5 6 7 8 9 10
1 2 4 8 6 9 5 12 20 7
87
Fixing the heap (from the top)

If we remove an element, we'll move each element
down one position in the array

2
4
8
6
9
5
12
20
7
0 1 2 3 4 5 6 7 8 9 10
2 4 8 6 9 5 12 20 7
88
Fixing the heap (from the top)

Then we'll have to reheap from the top down

2
4
8
6
9
5
12
20
7
89
Fixing the heap (from the top)

Now it's fixed
reheaping is O(log n)
Insertion or deletion is O(1), but then you have
to reheap
A Heap can be used to sort it's called Heap Sort

2
4
5
6
9
8
12
20
7
90
The PriorityQueue interface

public interface PriorityQueue
boolean isEmpty()
void add(Object x)
Object removeMin()
Object peekMin()

91
What is the output?

import ap.
public class PriorityQueueDemo
public static void main(String args)
PriorityQueue pq
new ArrayPriorityQueue()
pq.add(new Integer(5))
pq.add(new Integer(2))
pq.add(new Integer(3))
while(!pq.isEmpty())
System.out.println(pq.removeMin())

92
Collections

A collection is any bunch of objects you can
think of
For the AP AB exam, you are expected to know
Three interfaces Set, List and Map
Six classes
ArrayList, LinkedList, (implement List)
HashSet (implements Set)
TreeSet(implements SortedSet)
HashMap (implements Map)
TreeMap (implements SortedMap)

93
Collections and Iterators

An Iterator is like a loop
It allows you to go through the entire collection
in proper sequence
The three methods of the Iterator interface are
next, hasNext and remove

94
ArrayList vs. LinkedList

The code to use both is nearly identical
The difference is in the implementation
ArrayList uses an array, LinkedList is a linked
list
For large amounts of data, there may be
differences in performance
Most of the time, though, it really doesn't make
much difference which you use

95
ArrayList vs. LinkedList

Adding or deleting the front is O(n) for
ArrayList, O(1) for LinkedList
Accessing or changing the middle is O(1) for
ArrayList, O(n) for LinkedList
Inserting or deleting in the middle is O(n) for
both
In "real life", because the code is generic, it's
easy to test which is faster for a particular set
of data

96
Using an Iterator

import java.util.
import java.util.List
public class IteratorDemo
public static void main(String args)
List L new ArrayList()
L.add(new Integer(15))
L.add(new Integer(2))
L.add(new Integer(37))
Iterator itr L.iterator()
while(itr.hasNext())
System.out.println(itr.next())

97
Using an Iterator

import java.util.
import java.util.List
public class IteratorDemo
public static void main(String args)
List L new ArrayList()
L.add(new Integer(15))
L.add(new Integer(2))
L.add(new Integer(37))
for(Iterator itr L.iterator()
itr.hasNext())
System.out.println(itr.next())

98
The syntax for LinkedList is identical to
ArrayList

import java.util.
import java.util.List
public class IteratorDemo
public static void main(String args)
List L new LinkedList()
L.add(new Integer(15))
L.add(new Integer(2))
L.add(new Integer(37))
for(Iterator itr L.iterator()
itr.hasNext())
System.out.println(itr.next())

99
What is the output?

List L new LinkedList()
L.add(new Integer(15))
L.add(new Integer(2))
L.add(new Integer(37))
Iterator itr L.iterator()
while(itr.hasNext())
System.out.println(itr.next())
itr L.iterator()
while(itr.hasNext())
if(((Integer)itr.next()).intValue()5 0)
itr.remove()
itr L.iterator()
while(itr.hasNext())
System.out.println(itr.next())

100
Iterator vs. ListIterator

A ListIterator is exactly like an Iterator but
with two additional methods
void add(Object o) (adds to list before next
element)
void set(Object o) (replaces the last element
returned by next)

101
What is the output?

List l new LinkedList()
l.add(new String("Alices"))
l.add(new String("Adventures"))
l.add(new String("In"))
l.add(new String("Wonderland"))
ListIterator i l.listIterator()
while(i.hasNext())
String temp (String)i.next()
if(temp.length()2)
i.set(temp.toUpperCase())
i l.listIterator()
while(i.hasNext())
System.out.println(i.next())

102
What is the output?

List L new LinkedList()
L.add(new Integer(15))
L.add(new Integer(2))
L.add(new Integer(37))
ListIterator itr L.listIterator()
while(itr.hasNext())
if(((Integer)itr.next()).intValue()5 0)
itr.add(new Integer(3))
else
itr.set(new Integer(7))
Iterator itr2 L.iterator()
while(itr2.hasNext())
System.out.println(itr2.next())

103
hash tables

A hash table is an improvement of a bucket sort
less wasted space
can store 2 or more items in the same "bucket"
A hash table can be thought of as an array of
linked lists

104
hash tables

Let's say I have five numbers, each of which is
between 0 and 400
7 266 399 12 125
I don't want to make 400 "buckets", because 395
of them will be empty!
I'll take the number 5 and use that to decide
where to store it
I'm using 5 because there are 5 numbers to be
sortedto 5 buckets
In this example, 5 is the "hash function"

105
hash tables
7
0
266
1
399
2
12
3
125
4
106
hash tables
7 5 2
0
266
1
399
2
12
3
125
4
107
hash tables
0
266 5 1
1
399
2
7
12
3
125
4
108
hash tables
0
1
266
399 5 4
2
7
12
3
125
4
109
hash tables
0
1
266
2
7
12 5 2
3
125
4
399
110
hash tables
12 and 7 are placed in the location by the hash
function. This is called a collision. 12 is
inserted at the head of the linked list.
0
1
266
2
7
12
3
125
4
399
111
hash tables
0
1
266
2
7
12
3
125 5 0
4
399
112
hash tables
0
In this example, one of the five spots is
empty. of null pointers 20 The average
length of the (non-empty) linked lists are (1
1 2 1)/4 1.25 Ideally, we'd like to have 0
null pointers and an average length of 1. We
could write a different hash function and see if
it makes an improvement
125
1
266
2
7
12
3
4
399
113
hash function add the digits up and then 5
7 7 5 2
0
266 14 5 4
1
399
399 21 5 1
2
7
12 3 5 3
3
12
125
125 8 5 3
4
266
114
hash function sin(number) 2.5 2.5
sin(7)2.52.52
0
266
sin(266)2.52.50
1
sin(399)2.52.54
2
7
sin(12)2.52.53
3
12
sin(125)2.52.54
4
399
125
115
hash function .number 6
.764
0
125
12
.26661
1
266
.39962
2
399
.1260
3
.12560
4
7
116
hash tables

If there are no collisions, putting data in a
Hash table is O(1)
If there are no collisions, searching for a
number is O(1)
Just use the "hash function" to find where the
number would be stored

117
searching hash tables
0
Let's say I'm searching for the value 532. I do
5325 2. That means that 532 would be at
position 2 if it is in the hash table. We go to
the the linked list in position 2 and do a linear
search for the value 532.
125
1
266
2
7
12
3
4
399
118
hashCode()

You don't have to think of your own hashing
method
Every class inherits hashCode() from Object
Uses "Some Formula" we don't know or care about
Every Object is associated with an integeer value
called its hash code
equal objects have equal hash codes

119
The Set interface

No duplicate elements
Like a "mathematical" set
Implemented by HashSet and TreeSet
public interface Set
boolean add((Object obj)
boolean contains(Object obj)
boolean remove(Object obj)
int size()
Iterator iterator()

120
Using a HashSet to store a collection of people

import java.util.
public class HashSetDemo
public static void main(String args)
Set s new HashSet()
s.add("Mary")
s.add("Joan")
s.add("Mary") //note duplicate
s.add("Dennis")
System.out.println(s.size())
Iterator itr s.iterator()
while(itr.hasNext())
System.out.println(itr.next())
/ Sample Output
3

121
Using a TreeSet to store a collection of people

import java.util.
public class TreeSetDemo
public static void main(String args)
Set s new TreeSet()
s.add("Mary")
s.add("Joan")
s.add("Mary") //note duplicate
s.add("Dennis")
System.out.println(s.size())
Iterator itr s.iterator()
while(itr.hasNext())
System.out.println(itr.next())
/ Sample Outputnote different order of output
3

122
Comparing sets

import java.util.
public class SetDemo
public static void main(String args)
Set s new TreeSet()
s.add("Mary")
s.add("Joan")
s.add("Mary") //note duplicate
s.add("Dennis")
Set t new HashSet()
t.add("Mary")
t.add("Joan")
t.add("Dennis")
System.out.println(s.equals(t))

123
TreeSet vs. HashSet

HashSet is faster, but unordered
HashSet is O(1) for add, remove and contains
Iterating through a HashSet gives items in no
particular order
TreeSet is slower, but sorted
TreeSet is O(log n) for add, remove and contains
TreeSet is a balanced Binary Search Tree
TreeSet uses compareTo, so any items in the set
need to be "mutually comparable"
TreeSet is sorted in ascending order
Iterating through TreeSet gives all items in
ascending order

124
The Map interface

Implemented by HashMap and TreeMap
Each item must have unique "key"
No iterator! (e.g. you won't be expected to
display all items)
public interface Map
Object put(Object key, Object value)
Object get(Object key)
Object remove(Object key)
boolean containsKey(Object key)
int size()
Set keySet()

125
Using a HashMap to store a collection of people
using SS key

import java.util.
public class HashDemo
public static void main(String args)
Map h new HashMap()
h.put("123456789", "Joe Smith")
h.put("456719812", "Joe Smith")
h.put("731851382", "Homer Simpson")
h.put("725248225", "Joe Montana")
h.put("005049825", "Lisa Carlisle")
System.out.println(h.get("456719812"))
System.out.println(h.get("725248225"))

126
Using a TreeMap is identical

import java.util.
public class HashDemo
public static void main(String args)
Map h new TreeMap()
h.put("123456789", "Joe Smith")
h.put("456719812", "Joe Smith")
h.put("731851382", "Homer Simpson")
h.put("725248225", "Joe Montana")
h.put("005049825", "Lisa Carlisle")
System.out.println(h.get("456719812"))
System.out.println(h.get("725248225"))

127
Sets vs. Maps

Map has a get() method, Set doesn't
Maps can have duplicate items, as long as they
have different keys
Set has an iterator, Map doesn't
If you want to be able to retrieve a single item,
use a Map
If you will only deal with the items as a group,
use a Set

128
AP exam format

Both A and AB are two sections
40 multiple choice questions 115
4 free response questions 145
Usually, you are asked write a function for the
free response questions given a header and a task
One free response question will be about the MBCS
(the same question is usually on both exams)
5 or 6 multiple choice questions will be about
the MBCS
You will be given a "quick reference guide" to
the AP C classes and MBCS code

129
AP exam grading

Here are the statistics for the 1999 exam
Grade A exam AB exam
5 75 70
4 56 60
3 41 41
2 30 31
1 0 0

130
Ap exam tips

A exam one
"Distributing" the !
!(true false)
//is the same as
!true !false
//is the same as
false true
//is the same as
false

131
The worst mistake

Never display what you should return!
Problem Write a method that returns the sum of
two numbers
double Sum(double dOne, double dTwo)
System.out.println(dOne dTwo)
NO NO NO NO!

132
The worst mistake

Never display anything unless you are asked!
Problem Write a method that returns the sum of
two numbers
double Sum(double dOne, double dTwo)
System.out.println(dOne dTwo)
return dOne dTwo
NO NO NO NO!

133
The worst mistake

Now it's correct!
Problem Write a method that returns the sum of
two numbers
double Sum(double dOne, double dTwo)
return dOne dTwo
Beginning programmers typically want to display
everything. If they can't see it, they don't
believe it exists. Don't let the grader think
you're a rookie!

134
Design questions