Algorithms in Java, 4th Edition · Robert Sedgewick and Kevin Wayne · Copyright © 2009 · October 10, 2009 10:19:26 AM

4.2 Binary Search Trees

‣ BSTs ‣ ordered operations‣ deletion

Definition. A BST is a binary tree in symmetric order.

A binary tree is either:

• Empty.

• Two disjoint binary trees (left and right).

Symmetric order. Each node has a key, and every node’s key is:

• Larger than all keys in its left subtree.

• Smaller than all keys in its right subtree.

2

Binary search trees

right childof root

a left link

a subtree

root

null links

Anatomy of a binary tree

valueassociated

with R

parent of A and R

left linkof E

keys smaller than E keys larger than E

key

AC

E

HR

SX

9

Anatomy of a binary search tree

Java definition. A BST is a reference to a root Node.

A Node is comprised of four fields:

• A Key and a Value.

• A reference to the left and right subtree.

3

BST representation in Java

smaller keys larger keys

private class Node{ private Key key; private Value val; private Node left, right; public Node(Key key, Value val) { this.key = key; this.val = val; }}

Key and Value are generic types; Key is Comparable

Binary search tree

BST with smaller keys BST with larger keys

key

left right

val

BST

Node

public class BST<Key extends Comparable<Key>, Value>{ private Node root;

private class Node { /* see previous slide */ } public void put(Key key, Value val) { /* see next slides */ }

public Value get(Key key) { /* see next slides */ }

public void delete(Key key) { /* see next slides */ }

public Iterable<Key> iterator() { /* see next slides */ }

}

4

BST implementation (skeleton)

root of BST

Get. Return value corresponding to given key, or null if no such key.

5

BST search

R is less than Sso look to the left

black nodes couldmatch the search key

gray nodes cannotmatch the search key

found R(search hit)

so return value

R is greater than Eso look to the right

AC

E

HM

R

SX

AC

E

HM

R

SX

AC

E

HM

R

SX

T is less than Xso look to the left

link is nullso T is not in tree

(search miss)

T is greater than Sso look to the right

AC

E

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R

SX

AC

E

HM

R

SX

Successful (left) and unsuccessful (right ) search in a BST

successful search for R unsuccessful search for T

Get. Return value corresponding to given key, or null if no such key.

Running time. Proportional to depth of node.

6

BST search: Java implementation

public Value get(Key key) { Node x = root; while (x != null) { int cmp = key.compareTo(x.key); if (cmp < 0) x = x.left; else if (cmp > 0) x = x.right; else if (cmp == 0) return x.val; } return null; }

Put. Associate value with key.

Search for key, then two cases:

• Key in tree ⇒ reset value.

• Key not in tree ⇒ add new node.

7

BST insert

search for L endsat this null link

reset linkson the way up

create new node

AC

E

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AC

E

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L

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Insertion into a BST

inserting LPut. Associate value with key.

Running time. Proportional to depth of node.8

BST insert: Java implementation

public void put(Key key, Value val) { root = put(root, key, val); }

private Node put(Node x, Key key, Value val) { if (x == null) return new Node(key, val); int cmp = key.compareTo(x.key); if (cmp < 0) x.left = put(x.left, key, val); else if (cmp > 0) x.right = put(x.right, key, val); else if (cmp == 0) x.val = val; return x; }

concise, but tricky, recursive code;read carefully!

9

BST trace: standard indexing client

S

AC

E

HR

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AC

E

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S

AC

E

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AC

ER

S

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AE

S

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ES

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6

S 0

E 1

A 2

R 3

C 4

H 5

E 6

X 7

red nodesare new

black nodesare accessed

in search

changedvalue

changedvalue

changedvalue

gray nodesare untouched

AC

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BST trace for standard indexing client

key value key value • Many BSTs correspond to same set of keys.

• Cost of search/insert is proportional to depth of node.

Remark. Tree shape depends on order of insertion.

10

Tree shape

A

H

SR

X

CE

XS

RC

E

H

A

AC

E

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BST possibilities

best case

typical case

worst case

A

H

SR

X

CE

XS

RC

E

H

A

AC

E

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BST possibilities

best case

typical case

worst case

A

H

SR

X

CE

XS

RC

E

H

A

AC

E

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BST possibilities

best case

typical case

worst case

Observation. If keys inserted in random order, tree stays relatively flat.

11

BST insertion: random order

Typical BST built from random keys (N = 256)

12

BST insertion: random order visualization

Ex. Insert keys in random order.

13

Correspondence between BSTs and quicksort partitioning

Remark. Correspondence is 1-1 if no duplicate keys.

A

C

E

I

K

L

M

O

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Q

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T

U

UE

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BSTs: mathematical analysis

Proposition. If keys are inserted in random order, the expected number of compares for a search/insert is ~ 2 ln N.

Pf. 1-1 correspondence with quicksort partitioning.

Proposition. [Reed, 2003] If keys are inserted in random order,expected height of tree is ~ 4.311 ln N.

But… Worst-case for search/insert/height is N.(exponentially small chance when keys are inserted in random order)

15

ST implementations: summary

implementationguaranteeguarantee average caseaverage case ordered

ops?operations

on keysimplementation

search insert search hit insert ops? on keys

sequential search(unordered list) N N N/2 N no equals()

binary search(ordered array) lg N N lg N N/2 yes compareTo()

BST N N 1.39 lg N 1.39 lg N ? compareTo()

Costs for java FrequencyCounter 8 < tale.txt using BST

20

13

0

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‣ BSTs‣ ordered operations‣ deletion

Minimum. Smallest key in table.Maximum. Largest key in table.

Q. How to find the min / max.

Minimum and maximum

17

Examples of BST order queries

AC

E

HM

R

SX

min()max()max

min

Floor. Largest key ≤ to a given key.Ceiling. Smallest key ≥ to a given key.

Q. How to find the floor /ceiling.

Floor and ceiling

18

Examples of BST order queries

AC

E

HM

R

SX

min()max()

floor(D)

ceiling(Q)

floor(G)

Case 1. [k equals the key at root]The floor of k is k.

Case 2. [k is less than the key at root]The floor of k is in the left subtree.

Case 3. [k is greater than the key at root]The floor of k is in the right subtree(if there is any key ≤ k in right subtree);otherwise it is the key in the root.

Computing the floor

19

floor(G)in leftsubtree is null

result

!nding floor(G)

G is greater than E so floor(G) could be

on the right

G is less than S so floor(G) must be

on the left

AC

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Computing the "oor function

Computing the floor

20

floor(G)in leftsubtree is null

result

!nding floor(G)

G is greater than E so floor(G) could be

on the right

G is less than S so floor(G) must be

on the left

AC

E

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R

SX

AC

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Computing the "oor function

public Key floor(Key key){ Node x = floor(root, key); if (x == null) return null; return x.key;}private Node floor(Node x, Key key){ if (x == null) return null; int cmp = key.compareTo(x.key);

if (cmp == 0) return x;

if (cmp < 0) return floor(x.left, key);

Node t = floor(x.right, key); if (t != null) return t; else return x;

}

In each node, we store the number of nodes in the subtree rooted at that node.To implement size(), return the count at the root.

Remark. This facilitates efficient implementation of rank() and select().21

Subtree counts

A

A C E H M R S X

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A C E H M R S X

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2

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1

1

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1 1

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2

node count N

Two BSTs that representthe same set of keys

A

A C E H M R S X

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E

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A C E H M R S X

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2

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2

node count N

Two BSTs that representthe same set of keys

public int size() { return size(root); }

private int size(Node x) { if (x == null) return 0; return x.N; }

22

BST implementation: subtree counts

private class Node{ private Key key; private Value val; private Node left; private Node right; private int N;}

private Node put(Node x, Key key, Value val) { if (x == null) return new Node(key, val); int cmp = key.compareTo(x.key); if (cmp < 0) x.left = put(x.left, key, val); else if (cmp > 0) x.right = put(x.right, key, val); else if (cmp == 0) x.val = val; x.N = 1 + size(x.left) + size(x.right); return x; }

nodes in subtree

23

Rank

Rank. How many keys < k?

Easy recursive algorithm (4 cases!)

public int rank(Key key) { return rank(key, root); }

private int rank(Key key, Node x) { if (x == null) return 0; int cmp = key.compareTo(x.key); if (cmp < 0) return rank(key, x.left); else if (cmp > 0) return 1 + size(x.left) + rank(key, x.right); else return size(x.left); }

A

A C E H M R S X

C

E

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A

A C E H M R S X

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2

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2

node count N

Two BSTs that representthe same set of keys

• Traverse left subtree.

• Enqueue key.

• Traverse right subtree.

Property. Inorder traversal of a BST yields keys in ascending order.

key

key

val

BST with smaller keys

smaller keys, in order larger keys, in order

all keys, in order

BST with larger keys

left right

BST

Inorder traversal

24

public Iterable<Key> keys() { Queue<Key> q = new Queue<Key>(); inorder(root, queue); return q;}

private void inorder(Node x, Queue<Key> q) { if (x == null) return; inorder(x.left, q); q.enqueue(x.key); inorder(x.right, q); }

• Traverse left subtree.

• Enqueue key.

• Traverse right subtree.

Inorder traversal

25

function call stack

inorder(S) inorder(E) inorder(A) enqueue A inorder(C) enqueue C enqueue E inorder(R) inorder(H) enqueue H inorder(M) enqueue M print R enqueue S inorder(X) enqueue X

A C E H

M R S X

SS ES E A

S E A C

S E RS E R H

S E R H M

S X

queuerecursive calls

A

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BST: ordered symbol table operations summary

sequentialsearch

binarysearch

BST

search

insert

min / max

floor / ceiling

rank

select

ordered iteration

N lg N h

1 N h

N 1 h

N lg N h

N lg N h

N 1 h

N log N N N

h = height of BST(proportional to log N

if keys inserted in random order)

worst-case running time of ordered symbol table operations

27

‣ BSTs‣ ordered operations‣ deletion

28

ST implementations: summary

Next. Deletion in BSTs.

implementation

guaranteeguarantee average caseaverage caseaverage caseordered

iteration?operations

on keysimplementation

search insert delete search hit

insert deleteiteration? on keys

sequential search(linked list)

N N N N/2 N N/2 no equals()

binary search(ordered array)

lg N N N lg N N/2 N/2 yes compareTo()

BST N N N 1.39 lg N 1.39 lg N ??? yes compareTo()

29

BST deletion: lazy approach

To remove a node with a given key:

• Set its value to null.

• Leave key in tree to guide searches (but don't consider it equal to search key).

Cost. O(log N') per insert, search, and delete (if keys in random order),where N' is the number of key-value pairs ever inserted in the BST.

Unsatisfactory solution. Tombstone overload.

delete I

S

E

C

A

N

RH

I

S

E

C

A

N

RH

☠ tombstone

To delete the minimum key:

• Go left until finding a node with a null left link.

• Replace that node by its right link.

• Update subtree counts.

30

Deleting the minimum

public void deleteMin() { root = deleteMin(root); }

private Node deleteMin(Node x) { if (x.left == null) return x.right; x.left = deleteMin(x.left); x.N = 1 + size(x.left) + size(x.right); return x; }

go left untilreaching null

left link

return thatnode’s right link

available forgarbage collection

5

7

update links and countsafter recursive calls

AC

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Deleting the minimum in a BST

node to delete

replace withnull link

available forgarbage

collection

update counts afterrecursive calls

5

1

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deleting C

To delete a node with key k: search for node t containing key k.

Case 0. [0 children] Delete t by setting parent link to null.

31

Hibbard deletion

To delete a node with key k: search for node t containing key k.

Case 1. [1 child] Delete t by replacing parent link.

32

Hibbard deletion

node to deletereplace with

child link available forgarbage

collection

AC C C

E

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SX

AE

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deleting Rupdate counts after

recursive calls

5

7

To delete a node with key k: search for node t containing key k.

Case 2. [2 children]

• Find successor x of t.

• Delete the minimum in t's right subtree.

• Put x in t's spot.

33

Hibbard deletion

x has no left child

but don't garbage collect x

still a BST

search for key E

node to delete

deleteMin(t.right)

t

5

7

x

successor min(t.right)

t.left

x

update links andnode counts after

recursive calls

AC

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AC

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deleting E

Deletion in a BST

go right, thengo left until

reaching nullleft link

search for key E

node to delete

deleteMin(t.right)

t

5

7

x

successor min(t.right)

t.left

x

update links andnode counts after

recursive calls

AC

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deleting E

Deletion in a BST

go right, thengo left until

reaching nullleft link

34

Hibbard deletion: Java implementation

public void delete(Key key) { root = delete(root, key); }

private Node delete(Node x, Key key) { if (x == null) return null; int cmp = key.compareTo(x.key); if (cmp < 0) x.left = delete(x.left, key); else if (cmp > 0) x.right = delete(x.right, key); else { if (x.right == null) return x.left;

Node t = x; x = min(t.right); x.right = deleteMin(t.right); x.left = t.left; } x.N = size(x.left) + size(x.right) + 1; return x; }

no right child

replace with successor

search for key

update subtree counts

35

Hibbard deletion: analysis

Unsatisfactory solution. Not symmetric.

Surprising consequence. Trees not random (!) ⇒ sqrt(N) per op.Longstanding open problem. Simple and efficient delete for BSTs. Next lecture. Guarantee logarithmic performance for all operations.

36

ST implementations: summary

implementation

guaranteeguarantee average caseaverage caseaverage caseordered

iteration?operations

on keysimplementation

search insert deletesearch

hit insert deleteiteration? on keys

sequential search(linked list)

N N N N/2 N N/2 no equals()

binary search(ordered array)

lg N N N lg N N/2 N/2 yes compareTo()

BST N N N 1.39 lg N 1.39 lg N √N yes compareTo()

other operations also become √Nif deletions allowed