08BinarySearchTrees.txt

Binary Search Trees

(cid:1) basic implementations
(cid:1) randomized BSTs
(cid:1) deletion in BSTs

References:
    Algorithms in Java, Chapter 12
    Intro to Programming, Section 4.4
    http://www.cs.princeton.edu/introalgsds/43bst

1

Elementary implementations: summary

implementation

worst case

average case

search

insert

search

insert

ordered
iteration?

operations

on keys

unordered array

N

ordered array

lg N

unordered list

ordered list

N

N

N

N

N

N

N/2

N/2

lg N

N/2

N/2

N

N/2

N/2

no

yes

no

yes

equals()

compareTo()

equals()

compareTo()

Challenge:
     Efficient implementations of get() and put() and ordered iteration.

2

(cid:1) basic implementations
(cid:1) randomized BSTs
(cid:1) deletion in BSTs

3

Binary Search Trees (BSTs)

Def.  A BINARY SEARCH TREE is a binary tree in symmetric order.

A binary tree is either:
 empty
 a key-value pair and two binary trees
[neither of which contain that key]

it

best

was

the

of

times

equal keys ruled out to facilitate
associative array implementations

Symmetric order means that:
 every node has a key
 every nodes key is

larger than all keys in its left subtree
smaller than all keys in its right subtree

node

x

subtrees

smaller

larger

4

BST representation

A BST is a reference to a Node.

A Node is comprised of four fields:
 A key and a value.
 A reference to the left and right subtree.

smaller keys

larger keys

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

Key and Value are generic types;
Key is Comparable

root

it

2

best

1

was

2

the

1

of

1

times

1

5

BST implementation (skeleton)

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

instance variable

inner class

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

     
   public void put(Key key, Value val) 
   // see next slides

   public Val get(Key key)
   // see next slides
}

6

BST implementation (search)

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

root

it

2

best

1

was

2

the

1

get(the)

returns 1

of

1

times

1

get(worst)
returns null

7

BST implementation (insert)

public void put(Key key, Value val)

{  root = put(root, key, val);  }

root

best

1

it

2

was

2

put(the, 2)
overwrites the 1

the

1

worst

1

Caution: tricky recursive code.

Read carefully!

of

1

times

1

put(worst, 1)

adds a new entry

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.val = val;

   else if (cmp < 0) x.left  = put(x.left,  key, val);

   else if (cmp > 0) x.right = put(x.right, key, val);

   return x;

}

8

BST:  Construction

Insert the following keys into BST.    A S E R C H I N G X M P L 

9

Tree Shape

Tree shape.
 Many BSTs correspond to same input data.
 Cost of search/insert is proportional to depth of node.

typical

A

S

R

E

C

H

I

best

C

H

R

A

E

I

S

worst

A

C

E

Tree shape depends on order of insertion

H

I

R

S

10

BST implementation: iterator?

public Iterator<Key> iterator() 
{ return new BSTIterator(); }

private class BSTIterator
              implements Iterator<Key>
{
      
   BSTIterator()
   {   }

E

A

C

   public boolean hasNext()
   {   }

   public Key next()
   {   }
}

I

H

R

N

S

11

BST implementation: iterator?

Approach: mimic recursive inorder traversal

public void visit(Node x)
{
   if (x == null) return;
   visit(x.left)
   StdOut.println(x.key);
   visit(x.right);
}

E

A

C

Stack contents

I

H

R

N

visit(E)
  visit(A)
    print A
    visit(C)
      print C
  print E
  visit(S)
    visit(I)
      visit(H)
        print H
      print I
      visit(R)
        visit(N)
          print N
        print R
    print S

 
 A

 C
 E
  

 H

 I
 

 N
 R
 S

E
A  E
E
C  E
E

S
I  S
H  I  S
I  S
S
R  S
N  R  S
R  S
S

To process a node
 follow left links until empty

  (pushing onto stack)

 pop and process
 process node at right link

S

12

BST implementation: iterator

public Iterator<Key> iterator() 
{ return new BSTIterator(); }

private class BSTIterator
              implements Iterator<Key>
{
   private Stack<Node> 
             stack = new Stack<Node>();
   
   private void pushLeft(Node x)
   {
       while (x != null)
       {  stack.push(x); x = x.left; }
   }
   
   BSTIterator()
   {  pushLeft(root); }

   public boolean hasNext()
   {  return !stack.isEmpty(); }

   public Key next()
   {
      Node x = stack.pop();
      pushLeft(x.right);
      return x.key;
   }
}

E

A

C

I

H

R

N

   A  E 

A  C  E

C  E

E  H  I  S

H  I  S

I  N  R  S

N  R  S

R  S

S    

S

13

1-1 correspondence between BSTs and Quicksort partitioning

K

E

C

I

S

T

Q

A

E

M

R

X

L

P

O

no 
equal
 keys

U

14

BSTs:  analysis

Theorem.  If keys are inserted in random order, the expected number 
of comparisons for a search/insert is about 2 ln N.

(cid:1) 1.38 lg N, variance = O(1)

Proof: 1-1 correspondence with quicksort partitioning

Theorem.  If keys are inserted in random order, height of tree
is proportional to lg N, except with exponentially small probability.

mean (cid:1) 6.22 lg N, variance = O(1)

But...   Worst-case for search/insert/height is N.

e.g., keys inserted in ascending order

15

Searching challenge 3 (revisited):

Problem: Frequency counts in Tale of Two Cities

Assumptions: book has 135,000+ words
                      about 10,000 distinct words

Which searching method to use?
1) unordered array
2) unordered linked list
3) ordered array with binary search
4) need better method, all too slow
5) doesnt matter much, all fast enough
6) BSTs

insertion cost <  10000 * 1.38 * lg 10000 < .2 million
lookup cost < 135000 * 1.38 * lg 10000 <  2.5 million

16

Elementary implementations: summary

implementation

guarantee

average case

search

insert

search

insert

ordered
iteration?

operations

on keys

unordered array

N

ordered array

lg N

unordered list

ordered list

BST

N

N

N

N

N

N

N

N

N/2

lg N

N/2

N/2

N/2

N/2

N

N/2

1.38 lg N 1.38 lg N

no

yes

no

yes

yes

equals()

compareTo()

equals()

compareTo()

compareTo()

Next challenge:
   Guaranteed efficiency for get() and put() and ordered iteration.

17

(cid:1) basic implementations
(cid:1) randomized BSTs
(cid:1) deletion in BSTs

18

Rotation in BSTs

Two fundamental operations to rearrange nodes in a tree.
 maintain symmetric order.
 local transformations (change just 3 pointers).
 basis for advanced BST algorithms

Strategy: use rotations on insert to adjust tree shape to be more balanced

u

v

h

A

B

C

h = rotL(u)

h = rotR(v)

h

v

C

B

u

A

Key point: no change in search code (!)

19

Rotation

Fundamental operation to rearrange nodes in a tree.
 easier done than said
 raise some nodes, lowers some others

root = rotL(A)

private Node rotL(Node h)
{
   Node v = h.r;
   h.r = v.l;
   v.l = h;
   return v;
}

private Node rotR(Node h)
{
   Node u = h.l;
   h.l = u.r;
   u.r = h;
   return u;
}

A.left = rotR(S)

20

Recursive BST Root Insertion

insert G

Root insertion:  insert a node and make it the new root.
 Insert as in standard BST.
 Rotate inserted node to the root.
 Easy recursive implementation

Caution: very tricky recursive 

code.

Read very carefully!

private Node putRoot(Node x, Key key, Val val)

{

   if (x == null) return new Node(key, val);

   int cmp = key.compareTo(x.key);

   if (cmp == 0)  x.val = val;

   else if (cmp < 0)

   {  x.left  = putRoot(x.left,  key, val); x = rotR(x);  }

   else if (cmp > 0)

   {  x.right = putRoot(x.right, key, val); x = rotL(x);  }

   return x;

}

21

Constructing a BST with root insertion

Ex.   A S E R C H I N G X M P L 

Why bother?
 Recently inserted keys are near the top (better for some clients).
 Basis for advanced algorithms.

22

Randomized BSTs (Roura, 1996)

Intuition.  If tree is random, height is logarithmic. 
Fact.  Each node in a random tree is equally likely to be the root. 

Idea.  Since new node should be the root with probability 1/(N+1), 
make it the root (via root insertion) with probability 1/(N+1).

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.val = val; return x; }

   if (StdRandom.bernoulli(1.0 / (x.N + 1.0))

      return putRoot(h, key, val);

   if      (cmp < 0) x.left  = put(x.left,  key, val);

   else if (cmp > 0) x.right = put(x.right, key, val);

   x.N++;

   return x;

}

need to maintain count of 
nodes in tree rooted at x

23

Constructing a randomized BST

Ex:  Insert distinct keys in ascending order.

Surprising fact:
    Tree has same shape as if keys were
     inserted in random order.

Random trees result from any insert order

Note: to maintain associative array abstraction
need to check whether key is in table and replace
value without rotations if that is the case.

24

Randomized BST

Property.  Randomized BSTs have the same distribution as BSTs under 
random insertion order, no matter in what order keys are inserted.

 Expected height is ~6.22 lg N
 Average search cost is ~1.38 lg N.
 Exponentially small chance of bad balance.

Implementation cost.  Need to maintain subtree size in each node.

25

Summary of symbol-table implementations

implementation

guarantee

average case

search

insert

search

insert

ordered
iteration?

operations

on keys

unordered array

N

ordered array

lg N

unordered list

ordered list

BST

N

N

N

N

N

N

N

N

N/2

lg N

N/2

N/2

N/2

N/2

N

N/2

1.38 lg N 1.38 lg N

randomized BST

7 lg N 7 lg N 1.38 lg N 1.38 lg N

no

yes

no

yes

yes

yes

equals()

compareTo()

equals()

compareTo()

compareTo()

compareTo()

Randomized BSTs provide the desired guarantee

probabilistic, with
exponentially small

chance of quadratic time

Bonus (next): Randomized BSTs also support delete (!)

26

(cid:1) basic implementations
(cid:1) randomized BSTs
(cid:1) deletion in BSTs

27

BST delete: lazy approach

To remove a node with a given key
 set its value to null
 leave key in tree to guide searches

[but do not consider it equal to any search key]

E

A

C

I

H

R

N

S

E

remove  I

A

a tombstone

S

C

I

H

R

N

Cost. O(log N') per insert, search, and delete, where N' is the number 
of elements ever inserted in the BST.

Unsatisfactory solution:  Can get overloaded with tombstones.

28

BST delete: first approach

To remove a node from a BST.   [Hibbard, 1960s]
 Zero children:  just remove it.
 One child:  pass the child up.
 Two children: find the next largest node using right-left*

                      swap  with next largest
                      remove as above.

zero children 

one child

two children

Unsatisfactory solution.  Not symmetric, code is clumsy.
Surprising consequence.  Trees not random (!)  (cid:1) sqrt(N) per op.

Longstanding open problem: simple and efficient delete for BSTs

29

Deletion in randomized BSTs

To delete a node containing a given key
 remove the node
 join the two remaining subtrees to make a tree

Ex. Delete  S   in

E

A

C

S

X

I

H

R

N

30

Deletion in randomized BSTs

To delete a node containing a given key
 remove the node
 join its two subtrees

Ex. Delete  S  in

E

A

C

join these

two subtrees

X

I

H

R

N

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

31

Join in randomized BSTs

To join two subtrees with all keys in one less than all keys in the other
 maintain counts of nodes in subtrees (L and R)
 with probability L/(L+R)

make the root of the left the root
make its left subtree the left subtree of the root
join its right subtree to R to make the right subtree of the root

 with probability L/(L+R) do the symmetric moves on the right

to join these
two subtrees

I

H

R

N

make I the root

with probability 4/5

I

need to join these

two subtrees

X

H

X

R

N

32

Join in randomized BSTs

To join two subtrees with all keys in one less than all keys in the other
 maintain counts of nodes in subtrees (L and R)
 with probability L/(L+R)

make the root of the left the root
make its left subtree the left subtree of the root
join its right subtree to R to make the right subtree of the root

 with probability L/(L+R) do the symmetric moves on the right

private Node join(Node a, Node b)

{

   if (a == null) return a;
   if (b == null) return b;

   int cmp = key.compareTo(x.key);
   if (StdRandom.bernoulli((double)*a.N / (a.N + b.N))

     { a.right = join(a.right, b); return a; }

   else 
     { b.left  = join(a, b.left ); return b; }

}

to join these
two subtrees

X

R

N

make R the root

with probability 2/3

R

N

X

33

Deletion in randomized BSTs

To delete a node containing a given key
 remove the node
 join its two subtrees

Ex. Delete  S   in

E

A

C

I

H

R

N

E

S

A

I

X

C

H

R

N

X

Theorem.  Tree still random after delete (!)

Bottom line.  Logarithmic guarantee for search/insert/delete

34

Summary of symbol-table implementations

implementation

guarantee

average case

search insert delete

search

insert

delete

ordered
iteration?

unordered array

N

ordered array

lg N

unordered list

ordered list

BST

N

N

N

N

N

N

N

N

N

N

N

N

N

N/2

lg N

N/2

N/2

N/2

N/2

N

N/2

N/2

N/2

N/2

N/2

1.38 lg N 1.38 lg N

?

randomized BST 7 lg N 7 lg N 7 lg N 1.38 lg N 1.38 lg N 1.38 lg N

no

yes

no

yes

yes

yes

Randomized BSTs provide the desired guarantees

probabilistic, with
exponentially small

chance of error

Next lecture: Can we do better?

35