Binary Search Trees Briana B. Morrison Adapted from Alan Eugenio.

Binary Search Trees

Briana B. Morrison

Adapted from Alan Eugenio

Binary Trees 2

Topics

Applications Binary Search Trees

Retrieve Insert Delete

Binary Trees 3

Ordered Dictionaries

Keys are assumed to come from a total order.

New operations: closestBefore(k) closestAfter(k)

Binary Trees 4

Binary Search (§8.3.3)

Binary search performs operation find(k) on a dictionary implemented by means of an array-based sequence, sorted by key

similar to the high-low game at each step, the number of candidate items is halved terminates after O(log n) steps

Example: find(7)

1 3 4 5 7 8 9 11 14 16 18 19

1 3 4 5 7 8 9 11 14 16 18 19

1 3 4 5 7 8 9 11 14 16 18 19

1 3 4 5 7 8 9 11 14 16 18 19

0

0

0

0

ml h

ml h

ml h

lm h

Binary Trees 5

BINARY SEARCH TREES

Binary Trees 6

Overview of Binary Search TreeBinary search tree definition:

T is a binary search tree if either of these is true T is empty; or Root has two subtrees:

Each is a binary search tree Value in root > all values of the left subtree Value in root < all values in the right subtree

Binary Trees 7

HERE IS AN EXAMPLE OF A BINARY SEARCH TREE: 80

45 92

30 100 20 40 95 110 97

Binary Trees 8

50

30 90

12 40 86 100

25 60

71

Binary Trees 9

ANOTHER BINARY SEARCH TREE:

50

32

41

38

40

Binary Trees 10

Binary Search Trees

6 53 0

3 71 0

2 5

1 5

B in ary Search T ree 1


5 3

3 0

5 9

6 2

5 0


Binary Trees 11

A BINARY SEARCH TREE NEED NOT

BE FULL, COMPLETE OR A TWO-TREE,

BUT IT COULD BE ANY OF THOSE.

IF A BINARY SEARCH TREE IS FULL

OR COMPLETE, ITS HEIGHT IS

LOGARITHMIC IN n.

Binary Trees 12

IF A BINARY SEARCH TREE IS A

CHAIN, ITS HEIGHT IS LINEAR IN n.

EVEN BINARY SEARCH TREES THAT

ARE NOT CHAINS MAY HAVE HEIGHT

THAT IS LINEAR IN n. FOR EXAMPLE,

SUPPOSE THERE ARE EXACTLY TWO

ELEMENTS AT LEVEL 1, LEVEL 2, … .

SEE THE FOLLOWING TREE:

Binary Trees 13

80

45 92

30 100

95 110

93 97

Binary Trees 14

IN THE BinSearchTree CLASS, THE

“AVERAGE” HEIGHT OF A

BinSearchTree OBJECT IS

LOGARITHMIC IN n, AND SO THE

AVERAGE TIME TO INSERT, REMOVE

OR SEARCH IS LOGARITHMIC IN n.

Binary Trees 15

BUT A BinSearchTree OBJECT CAN BE

A CHAIN; THEN THE TIME TO INSERT,

REMOVE OR SEARCH IS LINEAR IN n.

Binary Trees 16

FOR AN (UNSORTED) ARRAY, vector,

deque, OR list CONTAINER, THE TIME

TO INSERT, REMOVE OR SEARCH IS

LINEAR IN n.

Binary Trees 17

THE BinSearchTree CLASS

Binary Trees 18

THE BinarySearchTree CLASS IS AN

ASSOCIATIVE-CONTAINER CLASS: THE

LOCATION OF AN ITEM IS

DETERMINED BY COMPARING THE

ITEM TO ITEMS ALREADY IN THETREE.

Binary Trees 19

THE BinSearchTree CLASS IS NOT PART

OF THE STANDARD TEMPLATE

LIBRARY, SO WE WILL TAKE A

PARSIMONIOUS APPROACH.

HERE ARE THE METHOD INTERFACES

FOR A MINIMAL BinarySearchTree

CLASS:

Binary Trees 20

1. // Postcondition: this BinSearchTree is empty.BinSearchTree( );

2. // Postcondition: the number of items in this// BinSearchTree has been returned.unsigned size( ) const;

3. // Postcondition: if there is an item in this BinSearchTree

// that equals item, the value returned is an// iterator pointing to that item. Otherwise,// the value returned is an iterator with the// same value as the iterator returned by// the end( ) method. The averageTime(n)// is O(log n) and worstTime(n) is O(n).Iterator find (const T& item) const;

Binary Trees 21

4. // Postcondition: item has been inserted into this// BinSearchTree. An iterator positioned at// the inserted item has been returned. The// averageTime(n) is O(log n) and// worstTime(n) is O(n).Iterator insert (const T& item);

5. // Precondition: itr is positioned at an item in this// BinSearchTree.// Postcondition: the item that itr is positioned at has been// deleted from this BinSearchTree. The// averageTime(n) is constant and// worstTime(n) is O(n). The// amortizedTime(n) is constant.void erase (iterator itr);

Binary Trees 22

6. // Postcondition: if this BinSearchTree is non-empty, an// iterator positioned at the smallest item// in the tree has been returned.// Otherwise, the iterator returned has// the same value as the iterator// returned by the end( ) method.Iterator begin( );

7. // Postcondition: the value returned is an iterator that// can be used in a comparison for// ending the traversal of this// BinSearchTree. If this BinSearchTree// is non-empty, the largest item is in the// position just before the position of the// iterator returned.Iterator end( );

Binary Trees 23

8. // Postcondition: The space allocated for this BinSearchTree// has been deallocated. The worstTime(n) is// O(n).~BinSearchTree( );

MISSING: push_back, push_front (BECAUSE USER CANNOT SAY WHERE TO PUT AN ITEM)

pop_front, pop_back (USE erase INSTEAD)

Binary Trees 24

EXERCISE: DETERMINE THE OUTPUT

FROM THE FOLLOWING CODE AFTER

THIS INPUT: summer spring winter maybe fall always ***

Binary Trees 25

BinSearchTree<string> words; BinSearchTree<string>::Iterator itr1; string word; cout << "Please enter a word; the sentinel is ***: "; cin >> word; while (word != "***") { words.insert (word); cout << "Please enter a word; the sentinel is ***: "; cin >> word; } // while for (itr1 = words.begin( ); itr1 != words.end( ); itr1++) cout << *itr1 << endl; words.erase (words.find ("maybe")); words.erase (words.begin( )); words.erase (--words.end( )); for (itr1 = words.begin( ); itr1 != words.end( ); itr1++) cout << *itr1 << endl;

summerspringwintermaybefallalways***

Binary Trees 26

Using Binary Search Trees Application: Removing Duplicates

7

52

37 3 2 5 3 2 9 3 2 3 5 7 9

9v v

Binary Trees 27

FIELDS AND ONE POSSIBLE IMPLEMENTATION OF THE

BinarySearchTree CLASS

Binary Trees 28

Binary Tree Nodes

Abs tract T re e M ode l

A

E F

H

D

CB

G

l e ft A ri gh t

T re e N ode M ode l

l e ft B ri gh t

l e ft E ri gh t

l e ft G ri gh t

l e ft D ri gh t

l e ft C ri gh t

l e ft H ri gh t

l e ft F ri gh t

Binary Trees 29

Searching a Binary Tree: Algorithm1. if root is NULL

2. item not in tree: return NULL3. compare target and root->data4. if they are equal5. target is found, return root->data6. else if target < root->data7. return search(left subtree)8. else9. return search(right subtree)

Note that in the STL, trees are not implemented with recursion but with loops

Binary Trees 30

50

62373210

60533525

5530

15

Current Node Action --LOCATING DATA IN A TREE-Root = 50 Compare item = 37 and 50

37 < 50, move to the left subtreeNode = 30 Compare item = 37 and 30

37 > 30, move to the right subtreeNode = 35 Compare item = 37 and 35

37 > 35, move to the right subtreeNode = 37 Compare item = 37 and 37. Item found.

Binary Trees 31

80

40 90

60

50 75

WHAT ARE THE STEPS IF THE CALL IS find (60)?

Binary Trees 32

80

40 90

60

50 75

WHAT ARE THE STEPS IF THE CALL IS find (70)?

Binary Trees 33

FOR A SUCCESSFUL SEARCH:

THE AVERAGE HEIGHT OF A BINARY SEARCH

TREE IS LOGARITHMIC IN n. SO averageTime(n) IS

O(log n). IN FACT, averageTime(n) IS LOGARITHMIC

IN n.

THE worstTime(n) OCCURS IF THETREE IS A

CHAIN. SO worstTime(n) IS ????

Binary Trees 34

Logic for Insert

Insertion is similar to find, but you must keep track of parent to be able to insert a new node

Find the spot in the tree that the value would be at if it were already in the tree,

When you reach a null link, insert the value there

Binary Trees 35

Update Operations: 1st of 3 steps1)- The function begins at the root node and compares item 32

with the root value 25. Since 32 > 25, we traverse the right subtree and look at node 35.

2 5

4 01 2

3 52 0

p aren t

t

(a)St ep 1 : C o m p are 3 2 an d 2 5 .T rav ers e t h e righ t s u b t ree.

Binary Trees 36

Update Operations: 2nd of 3 steps2)- Considering 35 to be the root of its own subtree, we compare

item 32 with 35 and traverse the left subtree of 35.

(b )St ep 2 : C o m p are 3 2 an d 3 5 .T rav ers e t h e left s u b t ree.

2 5

4 01 2

3 52 0

t

p aren t

Binary Trees 37

Update Operations: 3rd of 3 steps3)- Create a leaf node with data value 32. Insert the new node

as the left child of node 35.

newNode = getSTNode(item,NULL,NULL,parent);

parent->left = newNode;

(c)St ep 3 : In s ert 3 2 as left ch ildo f p aren t 3 5

2 5

4 01 2

3 52 0 p aren t

3 2

Binary Trees 38

insert (73) 80

40 90

60

50 75

Binary Trees 39

80

40 90

60

50 75

73

WILL THE INSERTED ITEM ALWAYS BE A LEAF?

Binary Trees 40

FOR INSERTING AN ITEM, WHAT IS THE WORST CASE? WHAT IS THE WORST HEIGHT?

THE worstTime(n) IS LINEAR IN n.

WHAT IS THE AVERAGE HEIGHT? THE averageTime(n) IS LOGARITHMIC IN n.

Binary Trees 41

WE NOW TACKLE THE MOST

DIFFICULT METHOD:

void erase (Iterator itr);

Binary Trees 42

TO DELETE THE NODE THAT itr IS

POSITIONED AT, WE NEED TO MAKE

AN ADJUSTMENT TO THE PARENT OF

THAT NODE.

Binary Trees 43

Deletion

There are three possible cases to deletion: Value to be deleted is a leaf node: just adjust

parent link and delete node Value to be deleted has one child: child

takes parent’s place (value to be deleted’s parent now points to grandchild) and delete node

Value to be deleted has two children: swap and delete node

Binary Trees 44

IF THE NODE THAT link IS POINTING TO

HAS NO CHILDREN OR ONE CHILD,

WE SIMPLY PRUNE THAT NODE FROM

THE TREE. FOR EXAMPLE,

Binary Trees 45

SUPPOSE link IS POINTING TO THE NODE WITH 50.

80

40 90

60

50 75

73

Binary Trees 46

80

40 90

60

75

73

Binary Trees 47

Removing an Item From a Binary Tree

1 0D

P

D e le te le a f n o d e 1 0 .p N o d e P tr-> le ft is d N o d e

N o re p la c e m e n t is n e c e s s a ry.p N o d e P tr-> le ft is N U L L

B efo re A ft er

4 0

3 5

6 53 0

5 0

3 32 6

2 5

3 42 9

2 8

P

4 0

3 5

6 53 0

5 0

3 32 6

2 5

3 42 9

2 8

Binary Trees 48

WHAT IF link IS POINTING TO THE NODE WITH 40?

80

40 90

60

75

73

Binary Trees 49

80

60 90

75

73

Binary Trees 50


D

P

D e le te n o de 3 5 with o n ly a le f t ch ild: No de R is th e le f t ch ild.

B e fo re

R

40

35

6530

50

3310 26

25

3429

28

P

A tta ch n o de R to th e pa re n t .

A fte r

R

40

33

6530

50

10 26

25

34

29

28

Binary Trees 51


P

D e le te n o d e 2 6 w ith o n ly a rig h t c h ild : N o d e R is th e rig h t c h ild .

P

A tta c h n o d e R to th e p a re n t.

B efo re A ft er

R

R

4 0

3 5

6 53 0

5 0

3 31 0 2 6

2 5

3 42 9

2 8

4 0

3 5

6 53 0

5 0

3 31 0 2 9

2 5

3 42 8

D

Binary Trees 52

WHAT IF THE NODE HAS TWO

CHILDREN?

Binary Trees 53


4 0

3 5

6 53 0

D e le te n o d e 3 0 w ith tw o c h ild re n .

5 0

3 31 0 2 6

2 5

3 42 9

2 8

4 0

3 5

6 5

O rp h a n e d s u b tre e s .

5 0

3 31 0

2 5

3 4

2 6

2 9

2 8

Binary Trees 54

Removing a Binary Search Tree Node

6 53 0

3 71 0

2 5

1 5

\\ //

D elet e n o d e 2 5

3 0

6 51 0

3 7

1 56 5

3 71 0

3 0

1 5

B ad So lu t io n : 3 0 is o u t o f p lace G o o d So lu t io n

(a) (b )

Binary Trees 55

SUPPOSE link IS POINTING TO 80’S NODE:

80

60 110

75 100 150

73 85 105

95

Binary Trees 56

THE ITEM 80 HAS TWO CHILDREN, SO WE

CANNOT SIMPLY UNLINK 80 FROM THE TREE:

THAT WOULD CREATE A HOLE.

OF THE ITEMS ALREADY IN THE TREE, WHICH

TWO COULD REPLACE 80 WITHOUT

DESTROYING THE STRUCTURE OF THE TREE?

Binary Trees 57

SUPPOSE link IS POINTING TO 80’S NODE:

80

60 110

75 100 150

73 85 105

95

Binary Trees 58

WE CAN REPLACE 80 WITH EITHER ITS PREDECESSOR, 75, OR ITS SUCCESSOR, 85.

WE’LL CHOOSE ITS SUCCESSOR. THE SUCCESSOR OF AN ITEM IS THE LEFTMOST ITEM IN THE RIGHT SUBTREE.

REPLACE 80 WITH 85, AND THEN REMOVE 85.

Binary Trees 59

HERE IS THE RESULTING TREE:

85

60 110

75 100 150

73 95 105

Binary Trees 60

CAN REMOVING THE SUCCESSOR GET

COMPLICATED? CAN THE SUCCESSOR

HAVE TWO CHILDREN?

Binary Trees 61


P

D

R

p O fR N o d eP t r = d N o d eP t r

rN o d eP t r

p N o d eP t r

4 0

3 5

6 53 0

5 0

3 31 0 2 6

2 5

3 42 9

2 8

P

R

R

4 0

3 5

6 53 0

5 0

3 31 0

2 6

3 4

2 9

2 8

B efo re rep lacin g D b y R A ft er rep lacin g D b y R

Binary Trees 62


D

R

d N o d eP t r

rN o d eP t rp O fR N o d eP t r

Pp N o d eP t r 4 0

3 5

6 53 0

5 0

3 31 0 2 6

2 5

3 42 9

2 8

P

R

4 0

3 5

6 53 0

5 0

3 41 0

2 6

3 3

2 9

2 8

p N o d eP t r

DR

rN o d eP t r

p O fR N o d eP t r

B efo re u n lin k in g R A ft er u n lin k in g R

Binary Trees 63


P

4 0

3 5

6 53 3

5 0

3 41 0

2 6

2 9

2 8

p N o d eP t r

RrN o d eP t r

P

4 0

3 5

6 53 0

5 0

3 41 0

2 6

2 9

2 8

p N o d eP t r

D3 3R

rN o d eP t r

B efo re rep lacin g D b y R A ft er rep lacin g D b y R

Binary Trees 64

Removing from a Binary Search Tree Item not present: do nothing

Item present in leaf: remove leaf (change to null) Item in non-leaf with one child:

Replace current node with that child Item in non-leaf with two children?

Find largest item in the left subtree Recursively remove it Use it as the parent of the two subtrees (Could use smallest item in right subtree)

Binary Trees 65

Where is itr++ positioned if itr is positioned at 61? At 59? At 36?

50 37 75 25 61 15 30 55 68 28 32 59 36

Binary Trees 66

RECALL THAT link IS THE ONLY FIELD IN THE Iterator

CLASS.

if (link has a right child)

// make link point to the leftmost node in link’s right subtreeelse // go up the tree to the left as far as possible, then go up // to the right. Make link point to that ancestor.

Binary Trees 67

FOR operator++( ), worstTime(n) IS

LINEAR IN n IF itr IS POSITIONED AT

THE ROOT AND THE ONLY LEAF IS

THE SUCCESSOR OF THE ROOT.

Binary Trees 68

DURING AN ITERATION OF ALL ITEMS,

EACH NODE IS ACCESSED AT MOST 3

TIMES (FOR ITS LEFT SUBTREE, FOR

ITSELF, AND FOR THE SUCCESSOR OF

ITS RIGHT SUBTREE). SO

amortizedTime(n), AND THEREFORE

averageTime(n) IS CONSTANT.

Binary Trees 69

EXERCISE: Suppose itr is positioned at the header, and the BinSearchTree is as follows:

50

40 60

70 65

WHERE IS ++itr POSITIONED?

Binary Trees 7070

Summary Slide 1§- A binary search tree stores data by

value instead of position- It is an example of an associative container.

§- The simple rules“== return”

“< go left”

“> go right”

until finding a NULL subtree make it easy to build a binary search tree that does not allow duplicate values.

Binary Trees 71

50

30 90

12 40 100 94 THIS IS A BINARY SEARCH TREE: EACH ITEM IN THE

LEFT SUBTREE IS LESS THAN THE ROOT ITEM, EACH

ITEM IN THE RIGHT SUBTREE IS GREATER THAN

THE ROOT ITEM, AND THE LEFT AND RIGHT SUB-

TREES ARE THEMSELVES BINARY SEARCH TREES.

Binary Trees 7272

Summary Slide 2§- The insertion algorithm can be used to define the

path to locate a data value in the tree.

§- The removal of an item from a binary search tree is more difficult and involves finding a replacement node among the remaining values.

Binary Search Trees Briana B. Morrison Adapted from Alan Eugenio.

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Transcript of Binary Search Trees Briana B. Morrison Adapted from Alan Eugenio.