Trees (Revisited) CHAPTER 15 6/30/15 Nyhoff, ADTs, Data Structures and Problem Solving with C++, Second Edition, © 2005 Pearson Education, Inc. All rights

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ata Structures and P

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1

Trees(Revisited)

CHAPTER 15

6/30/15

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2Chapter Contents

15.1 Case Study: Huffman Codes

15.2 Tree Balancing: AVL Trees

15.3 2-3-4 TreesRed-Black TreesB-TreesOther Trees

15.4 Associative Containers in STL – Maps (optional)

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3Chapter Objectives

Show how binary trees can be used to develop efficient codes

Study problem of binary search trees becoming unbalanced Look at classical AVL approach for solution

Look at other kinds of trees, including 2-3-4 trees, red-black trees, and B-trees

Introduce the associate containers in STL and see how red-black trees are used in implementation

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4Case Study: Huffman Codes

Recall that ASCII, EBCDIC, and Unicode use same size data structure for all characters

Contrast Morse code Uses variable-length sequences

Some situations require this variable-length coding scheme Can save space vs set data size

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5Variable-Length Codes Each character in such a code

Has a weight (probability) and a length

Expected length: expected length of the code for any one “codeword”

The expected length is the sum of the products of the weights and lengths for all the characters

(0.2 x 2) + (0.1 x 4) + (0.1 x 4) + (0.15 x 3) + (0.45) x 1 = 2.1

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6Immediate Decodability When no sequence of bits that

represents a character is a prefix of a longer sequence for another character Can be decoded without waiting for remaining bits

Note how previous scheme is not immediately decodable

And this one isCharacter

Code

A 01

B 10

C 010

D 0000

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7Huffman Codes We seek codes that are

Immediately decodable Each character has minimal expected

code length

For a set of n characters { C1 .. Cn } with weights { w1 .. wn }

We need an algorithm which generates n bit strings representing the codes

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8Huffman's Algorithm

1. Initialize list of n one-node binary trees containing a weight for each character

2. Do the following n – 1 timesa. Find two trees T' and T" in list with minimal weights w' and w"b. Replace these two trees with a binary tree whose root is w' + w" and whose subtrees are T' and T"and label points to these subtrees 0 and 1

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9Huffman's Algorithm

3. The code for character Ci is the bit string labeling a path in the final binary tree form from the root to Ci

Given characters

The end with codes

result is

Character Huffman Code

A 011

B 000

C 001

D 010

E 1

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10Overall Result

Each time we traverse a left branch, we add a 0 to the code

Each time we traverse a right branch, we add a 1 to the code

These variable length codes, take advantage of the frequencies of the symbols This results in time and memory efficiencies

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11Huffman Decoding Algorithm1. Initialize pointer p to root of Huffman tree

2. While end of message string not reached do the following

a. Let x be next bit in stringb. if x = 0 set p equal to left child pointer else set p to right child pointerc. If p points to leaf

i. Display character with that leaf ii. Reset p to root of Huffman tree

12Huffman Decoding Algorithm

For message string 01010010110101 Using Hoffman Tree and decoding algorithm

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Click for answerClick for answer

010 1 001 011 010 1

D E C A D E

Initial Leaves:{ (A20) (B10) (C10) (D15) (E45) }

Merge { (A20) ({BC} 20) (D15) (E45) }

Merge { ( {AD}35 ) ( {BC}20 ) (E45) }

Merge { ( {ADBC}55 ) (E45) }

Merge { ( { ADBCE}100 ) }

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13Exercise #1: Huffman Tree

0

0

00 0

0

1

1

11

1

1

1

1. Create the Huffman tree following the algorithm. Take the two lowest weight nodes and merge them2. Label the vertices with the weight and letter3. Label the 0/14. Create a table with overall encoding

Character Huffman Code

A 0

B 100

C 1010

D 1011

E 1100

F 1101

G 1110

H 1111

0

A B C D E F G H

8 3 1 1 1 1 1 1

Ref: https://mitpress.mit.edu/sicp/full-text/sicp/book/node41.html

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14Tree Balancing: AVL Trees

Consider a BST of state abbreviations

When tree is nicely balanced, search time is O(log2n)

If abbreviations entered in orderDE, GA, IL, IN, MA, MI, NY, OH, PA, RI, TX, VT, WY BST degenerates into linked list with

search time O(n)

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15AVL Trees

A height balanced tree Named with initials of Russian mathematicians who

devised them

Georgii Maksimovich Adel’son-Vel’ski and Evgenii Mikhailovich Landis

Defined as Binary search tree

Balance factor of each node is 0, 1, or -1

(Balance factor of a node = left height of sub tree minus right height of subtree)

(Balance factor of a node = left height of sub tree minus right height of subtree)

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16AVL Trees

Consider linked structure to represent AVL tree nodes Includes data member for the balance

factor

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17AVL Trees Insertions may require adjustment of the tree to maintain

balance Given tree

Becomes unbalanced

Requires right rotation on subtree of RI

Producing balanced tree

Insert DEInsert DE

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18Basic Rebalancing Rotations

1. Simple right rotationWhen inserted item in left subtree of left child of nearest ancestor with factor +2

2. Simple left rotationWhen inserted item in right subtree of right child of nearest ancestor with factor -2

3. Left-right rotationWhen inserted item in right subtree of left child of nearest ancestor with factor +2

4. Right-left rotationWhen inserted item in left subtree of right child of nearest ancestor with factor -2

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Rotations carried out by resetting links

Right rotation with A = nearest ancestor of inserted item with balance factor +2 and B = left child Reset link from parent of A to B

Set left link of A equal to right link of B

Set right link of B to point A

Next slide shows this sequence

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Right-Left rotationItem C inserted in right subtree of left child B of nearest ancestor A

1. Set left link of A to point to root of C2. Set right link of B equal to left link of

C3. Set left link of C to point to B

Now the right rotation4. Reset link from parent of A to point

to C5. Set link of A equal to right link of C6. Set right link of C to point to A

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22Basic Rebalancing Rotations When B has no right child before node C insertion

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23Another way to look at rotations…

Exercise #2: Apply a Left-Rotate operation on node 15

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24Exercise #3: Create an AVL Tree

Create an AVL tree using the following numbers: 4, 8, 12, 1, 2, 3

Show each step/rotation

4

2

1 3

8

12

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25Steps 1-3:

4

8

12

4 Step 1: Initial tree created with value

4. By default this is balanced

Step 2: The node 8 is added which is greater than 4, so it becomes a right node. It by itself is balanced, but at the root, there is no left subtree. Left subtree = 0, Right subtree = 1 so root balance factor = -1

Step 3: The node 12 is added, which is greater than 4 and 8, so it becomes the right node of 4. It by itself is balanced, but 8 now has an unbalanced subtree because it does not have a left subtree. Now the root, 4, has no left subtree but a height of two on it’s right subtree, so it’s balance factor is 0 – 2 = -2

4

8

0

-1

0

-2

-1

0

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26Step 4:4

8

12

-2

-1

Step 4: Because we inserted a right child on the right subtree that caused the imbalance, we need to apply a left rotation. The imbalance occurs at the root (4) so therefore we rotate at node (8)

The result is a balanced tree with 8 as the new root

0

4

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27Step 5: Next we insert node (1) to the

left of 4. We now have an additional level of height in the left subtree, but no additional rotation is needed yet.

Next, we insert node (2). This node is less than 8, less than 4 and greater than 1, so it gets attached as the right child in the left subtree. As you can see we now have an imbalance which we will handle by a R-L rotation next.

4

8

12

1

1

01

0

4

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2

2

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-12

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28Step 6: The imbalance is at node (4),

so therefore the rotation will be at node (1). First we apply a left rotation

We can see after the left rotation, that there is still an imbalance, which is why we need to apply the right rotation at the location where node (2) now resides.

4

8

12

2

2

01

-12

0

4

8

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2

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02

11

0

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29Step 7: The imbalance is still

at node (4), and we need to complete our second step of our L-R rotation, so therefore the rotation will be at node (2). We apply the second step, the Right rotation.

We can now see that the tree is again balanced.

4

8

12

2

2

02

11

0

4

8

12

0

1

01

0

2

0

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On our last step, we insert node 3. This is less than 8, but greater than 2 and less than 4, so it becomes the left child of 4.

Left-right rotationWhen inserted item in right subtree of left child of nearest ancestor with factor +2

Note that the imbalance occurs at the root, and that the offending insertion was on the right subtree (4 – 3) of the left child (2) of the nearest ancestor with a factor of 2 (node 8). So therefore we need a L-R rotation with the rotation occurring at node(2).

We apply the first part, the left rotation

4

8

12

1

2

01

0

2

-1

30

3

0

3

8

12

0

2

02

0

4

2

1

0

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Finally we need to apply the last part, the “right” rotation (of the L-R rotation). This rotation is applied at node (4).

This is the last step. We end with a balanced tree.

30

3

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3

8

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-1

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32Non Binary Trees

Some applications require more than two children per node

Genealogical tree

Game tree

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332-3-4 Trees

We extend searching technique for BST At a given node, we now have multiple paths a

search path may follow

Define m-node in a search tree

Stores m – 1 data values k1 < k2 … < km-1

Has links to m subtrees T1 ..Tm

Where for each i all data values in Ti < ki ≤ all values in Tk+1

342-3-4 Trees Example of m-node

Define 2-3-4 tree Each node stores at most 3 data values

Each internode is a 2-node, a 3-node, or a 4-node

All the leaves are on the same level

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35What is a 2, 3 or 4 node?

A 2-node will have one data element and two separate child nodes Like a binary search tree

A 3-node will have two data elements and 3 separate child nodes

A 4-node will have three data elements (this is the max), and 4 separate child nodes

Note that the child nodes represent values in between the parent node values

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362-3-4 Trees

Example of a 2-3-4 node which stores integers

To search for 36 Start at root … 36 < 53, go left

Next node has 27 < 36 < 38. take middle link

Find 36 in next lowest node

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37Building a Balanced 2-3-4 Tree

If tree is empty Create a 2-node containing new item, root of

tree

Otherwise

1. Find leaf node where item should be inserted by repeatedly comparing item with values in node and following appropriate link to child

2. If room in leaf nodeAdd itemOtherwise … (ctd)

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Otherwise:

a. Split 4-node into 2 nodes, one storing items less than median value, other storing items greater than median. Median value moved to a parent having these 2 nodes as children

b. If parent has no room for median, spilt that 4-node in same manner, continuing until either a parent is found with room or reach full root

c. If full root 4-node split into two nodes, create new parent 2-node, which = the new root

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39Exercise #4: Building a Balanced 2-3-4 Tree

Take the following list of numbers and using the algorithm, construct a 2-3-4 tree

53, 27, 75, 25, 70, 41, 38, 16, 59, 36, 73, 65, 60, 46, 55, 33, 68, 79, 48

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Note that last step in algorithm may require multiple splits up the tree

Instead of bottom-up insertion Can do top-down insertion

Do not allow any parent nodes to become 4-nodes

Split 4-nodes alongsearch pathas encountered

Note that 2-3-4 trees stay balanced during

insertions and deletions

Note that 2-3-4 trees stay balanced during

insertions and deletions

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41Implementing 2-3-4 Trees

class Node234

{

public

DataType data[3]; // type of data item in nodes

Node234 * child[4];

// Node234 operations

};

typedef Node234 * Pointer234

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42Red-Black Trees

Defined as a BST which: Has two kinds of links (red and black)

Every path from root to a leaf node has same number of black links

No path from root to leaf node has more than two consecutive red links

Data member added to store color of link from parent

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43Red-Black Trees

enum ColorType {RED, BLACK};

class RedBlackTreeNode

{

public:

DataType data;

ColorType parentColor; // RED or BLACK

RedBlackTreeNode * parent;

RedBlackTreeNode * left;

RedBlackTreeNode * right;

}

Now possible to construct a red-black tree to represent a 2-3-4 tree

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44Red-Black Trees 2-3-4 tree represented by red-black trees as follows:

Make a link black if it is a link in the 2-3-4 tree

Make a link red if it connects nodes containing values in same node of the 2-3-4 tree

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45Red-Black Trees Example: given 2-3-4 tree

Represented by this red-black tree70

59 60 65 73 75

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46Constructing Red-Black Tree Do top-down insertion as with 2-

3-4 tree

1. Search for place to insert new node – Keep track of parent, grandparent, great grandparent

2. When 4-node q encountered, split as follows:a. Change both links of q to blackb. Change link from parent to red:

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47Constructing Red-Black Tree

3. If now two consecutive red links, (from

grandparent gp to parent p to q)Perform appropriate AVL type rotation determined by direction (left-left, right-right,

left-right, right-left) from gp -> p-> q

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48B-Trees

Previous trees used in internal searching schemes Tree sufficiently small to be all in memory

B-tree useful in external searching Data stored in 2ndry memory

B-tree has properties Each node stores at most m – 1 data values Each internal nodes is 2-node, 3-node, …or m-node All leaves on same level

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49B-Trees

Thus a 2-3-4 tree is a B-tree of order 4

Note example below of order 5 B-tree

Best performance found to be with values for 50 ≤ m ≤ 400

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50Representing Trees & Forests as Binary Trees

Consider a node with multiple links

const int MAX_CHILDREN = … ;

class TreeNode

{

public:

DataType data

TreeNode * child[MAX_CHILDREN];

// TreeNode operations

}

typedef TreeNode * TreePointer;

Note problem with wasted space

when not all links used

Note problem with wasted space

when not all links used

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Use binary (two link) tree to represent multiple links Use one of the links to connect siblings in

order from left to right The other link points to first node in this

linked list of its children Note right pointer in root always null

Note right pointer in root always null

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Now possible to represent collection of trees (a forest!)

This collection becomes

Documents

Trees (Revisited) CHAPTER 15 6/30/15 Nyhoff, ADTs, Data Structures and Problem Solving with C++, Second Edition, © 2005 Pearson Education, Inc. All rights