Trees (Ch. 10.2)

Longin Jan LateckiTemple University

based on slides bySimon Langley,

Shang-Hua Teng, and William Albritton

Section 10.2: Applications of Trees

Binary search trees• A simple data structure for sorted lists

Decision trees• Minimum comparisons in sorting algorithms

Prefix codes• Huffman coding

Basic Data Structures - Trees

Informal: a tree is a structure that looks like a real tree (up-side-down)

Formal: a tree is a connected graph with no cycles.

Trees - Terminology

x

b e m

c d a

root

leaf

height=2

size=7

Every node must have its value(s)Non-leaf node has subtree(s)Non-root node has a single parent node

value

subtree

nodes

Types of Tree

Binary Tree

m-ary Trees

Each node has at most 2 sub-trees

Each node has at most m sub-trees

Binary Search Trees

A binary search tree: … is a binary tree. if a node has value N, all values in its

left sub-tree are less than or equal to N, and all values in its right sub-tree are greater than N.

Binary Search Tree Format

Items are stored at individual tree nodes.

We arrange for the tree to always obey this invariant:

For every item x,• Every node in x’s left

subtree is less than x.

• Every node in x’s right subtree is greater than x.

7

3 12

1 5 9 15

0 2 8 11

Example:

This is NOT a binary search tree

5

4 7

3 2 8 9

This is a binary search tree

Searching a binary search tree

search(t, s) {

If(s == label(t))

return t;

If(t is leaf) return null

If(s < label(t))

search(t’s left tree, s)

else

search(t’s right tree, s)}

h

Time per level

O(1)

O(1)

Total O(h)

Searching a binary search tree

search( t, s )

{ while(t != null)

{ if(s == label(t)) return t;

if(s < label(t)

t = leftSubTree(t);

else

t = rightSubTree(t);

}

return null;

h

Time per level

O(1)

O(1)

Total O(h)

Here’s another function that does the same (we search for label s):

TreeSearch(t, s)

while (t != NULL and s != label[t])

if (s < label[t])

t = left[t];

else

t = right[t];

return t;

Insertion in a binary search tree:we need to search before we insert

5

3 8

2 4 7 9

Time complexity ?

Insert 6 6

6

6

6

Insert 1111

11

11

O(height_of_tree)O(log n) if it is balanced n = size of the tree

always insert to a leaf

Insertion

insertInOrder(t, s)

{ if(t is an empty tree) // insert here

return a new tree node with value s

else if( s < label(t))

t.left = insertInOrder(t.left, s )

else

t.right = insertInOrder(t.right, s)

return t }

Recursive Binary Tree Insert

procedure insert(T: binary tree, x: item)v := root[T]if v = null then begin

root[T] := x; return “Done” endelse if v = x return “Already present”else if x < v then

return insert(leftSubtree[T], x)else {must be x > v}

return insert(rightSubtree[T], x)

Comparison –Insertion in an ordered list

Insert 6

Time complexity?

2 3 4 5 7 98

6 6 6 6

6

O(n) n = size of the list

insertInOrder(list, s) { loop1: search from beginning of list, look for an item >= s loop2: shift remaining list to its right, start from the end of list insert s}

6 7 8 9

Try it!!

Build binary search trees for the following input sequences• 7, 4, 2, 6, 1, 3, 5, 7

• 7, 1, 2, 3, 4, 5, 6, 7

• 7, 4, 2, 1, 7, 3, 6, 5

• 1, 2, 3, 4, 5, 6, 7, 8

• 8, 7, 6, 5, 4, 3, 2, 1

Decision Trees

A decision tree represents a decision-making process.• Each possible “decision point” or situation is

represented by a node.

• Each possible choice that could be made at that decision point is represented by an edge to a child node.

In the extended decision trees used in decision analysis, we also include nodes that represent random events and their outcomes.

Coin-Weighing Problem

Imagine you have 8 coins, oneof which is a lighter counterfeit, and a free-beam balance.• No scale of weight markings

is required for this problem!

How many weighings are needed to guarantee that the counterfeit coin will be found?

?

As a Decision-Tree Problem In each situation, we pick two disjoint and

equal-size subsets of coins to put on the scale.

The balance then“decides” whether to tip left, tip right, or stay balanced.

A given sequence ofweighings thus yieldsa decision tree withbranching factor 3.

Applying the Tree Height Theorem

The decision tree must have at least 8 leaf nodes, since there are 8 possible outcomes.• In terms of which coin is the counterfeit one.

Recall the tree-height theorem, h≥logm.• Thus the decision tree must have height

h ≥ log38 = 1.893… = 2. Let’s see if we solve the problem with only 2

weightings…

General Solution Strategy The problem is an example of searching for 1 unique particular

item, from among a list of n otherwise identical items. • Somewhat analogous to the adage of “searching for a needle in

haystack.” Armed with our balance, we can attack the problem using a divide-

and-conquer strategy, like what’s done in binary search.• We want to narrow down the set of possible locations where the

desired item (coin) could be found down from n to just 1, in a logarithmic fashion.

Each weighing has 3 possible outcomes.• Thus, we should use it to partition the search space into 3 pieces that

are as close to equal-sized as possible. This strategy will lead to the minimum possible worst-case number

of weighings required.

Coin Balancing Decision Tree

Here’s what the tree looks like in our case:

123 vs 456

1 vs. 2

left:123 balanced:

78right:456

7 vs. 84 vs. 5

L:1 R:2 B:3 L:4R:5 B:6 L:7 R:8

General Balance Strategy

On each step, put n/3 of the n coins to be searched on each side of the scale.• If the scale tips to the left, then:

• The lightweight fake is in the right set of n/3 ≈ n/3 coins.

• If the scale tips to the right, then:• The lightweight fake is in the left set of n/3 ≈ n/3 coins.

• If the scale stays balanced, then:• The fake is in the remaining set of n − 2n/3≈ n/3 coins

that were not weighed!

You can prove that this strategy always leads to a balanced 3-ary tree.

Suppose we have 3GB character data file that we wish to include in an email.

Suppose file only contains 26 letters {a,…,z}. Suppose each letter in {a,…,z} occurs with frequency

f. Suppose we encode each letter by a binary code If we use a fixed length code, we need 5 bits for each

character The resulting message length is

Can we do better?

Data Compression

zba fff 5

Data Compression: A Smaller Example Suppose the file only has 6 letters {a,b,c,d,e,f}

with frequencies

Fixed length 3G=3000000000 bits Variable length

110011011111001010

101100011010001000

05.09.16.12.13.45.

fedcba

Fixed length

Variable length

G24.2405.409.316.312.313.145.

How to decode?

At first it is not obvious how decoding will happen, but this is possible if we use prefix codes

Prefix Codes No encoding of a

character can be the prefix of the longer encoding of another character:

We could not encode t as 01 and x as 01101 since 01 is a prefix of 01101

By using a binary tree representation we generate prefix codes with letters as leaves

e

a

t

n s

0 1

1

1

1

0

0

0

Decoding prefix codes

Follow the tree until it reaches to a leaf, and then repeat!

A message can be decoded uniquely!

Prefix codes allow easy decoding

e

a

t

n s

0 1

1

1

1

0

0

0

Decode:

11111011100

s 1011100

sa 11100

san 0

sane

Some Properties

Prefix codes allow easy decoding An optimal code must be a full binary tree (a

tree where every internal node has two children)

For C leaves there are C-1 internal nodes The number of bits to encode a file is

ccfT TCc

length )()B(

where f(c) is the freq of c, lengthT(c) is the tree depth of c, which corresponds to the code length of c

Optimal Prefix Coding Problem

Given is a set of n letters (c1,…, cn) with frequencies (f1,…, fn).

Construct a full binary tree T to define a prefix code that minimizes the average code length

iT

n

i i cfT length )Average(1

Greedy Algorithms

Many optimization problems can be solved using a greedy approach• The basic principle is that local optimal decisions may be used to

build an optimal solution

• But the greedy approach may not always lead to an optimal solution overall for all problems

• The key is knowing which problems will work with this approach and which will not

We study• The problem of generating Huffman codes

Greedy algorithms A greedy algorithm always makes the choice that looks

best at the moment• My everyday examples:

• Driving in Los Angeles, NY, or Boston for that matter

• Playing cards

• Invest on stocks

• Choose a university

• The hope: a locally optimal choice will lead to a globally optimal solution

• For some problems, it works

Greedy algorithms tend to be easier to code

David Huffman’s idea

A Term paper at MIT

Build the tree (code) bottom-up in a greedy fashion

Each tree has a weight in its root and symbols as its leaves.

We start with a forest of one vertex trees representing the input symbols.

We recursively merge two trees whose sum of weights is minimal until we have only one tree.

The Huffman Coding algorithm- History

In 1951, David Huffman and his MIT information theory classmates given the choice of a term paper or a final exam

Huffman hit upon the idea of using a frequency-sorted binary tree and quickly proved this method the most efficient.

In doing so, the student outdid his professor, who had worked with information theory inventor Claude Shannon to develop a similar code.

Huffman built the tree from the bottom up instead of from the top down

Huffman Coding Algorithm

1. Take the two least probable symbols in the alphabet

2. Combine these two symbols into a single symbol, and repeat.

Example

Ax={ a , b , c , d , e }

Px={0.25, 0.25, 0.2, 0.15, 0.15}

d0.15

e0.15

b0.25

c0.2

a0.25

0.3

0 1

0.45

0 1

0.55

0

1

1.0

0

1

00 10 11 010 011

Building the Encoding Tree

Building the Encoding Tree

Building the Encoding TreeBuilding the Encoding Tree

Building the Encoding TreeBuilding the Encoding Tree

Building the Encoding TreeBuilding the Encoding Tree