Presented By:Susheel Thakur

M.Tech(C.S)Roll No- 324

Have you ever thought how does the OS manages or files?

Why do we have a hierarchical file system?

How do files get saved and deleted under hierarchical directories ?

SOLUTION: A hierarchical data structure called Trees.

04/12/23 04:50 AM 2Data Structures and Algorithms

Tree Nodes Each node can have 0 or more children A node can have at most one parent

Binary tree Tree with 0–2 children per node

Tree Binary Tree04/12/23 04:50 AM 3Data Structures and Algorithms

Terminology Root no parent Leaf no child Interior non-leaf Height maximum level of a node Level Length of path from root to node plus

1 Root node

Leaf nodes

Interior nodes Height

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General Tree – organization chart format Parenthetical Listing

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Parenthetical Listing – the algebraic expression, where each open parenthesis indicates the start of a new level and each closing parenthesis completes the current level and moves up one level in the tree.

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A (B (C D) E F (G H I) )

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List Representation( A ( B ( E ( K, L ), F ), C ( G ), D ( H ( M ), I, J ) ) )The root comes first, followed by a list of sub-trees

data link 1 link 2 ... link n

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Every tree node: object – useful information children – pointers to its children nodes

O

O O

O

O

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A binary tree is a tree in which no node can have more than two subtrees; the maximum outdegree for a node is two.

In other words, a node can have zero, one, or two subtrees.

These subtrees are designated as the left subtree and the right subtree.

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A null tree is a tree with no nodes

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The height of binary trees can be mathematically predicted

Given that we need to store N nodes in a binary tree, the maximum height is

maxH NA tree with a maximum height is rare. It occurs when all of the nodes in the entire tree have only one successor.

14Data Structures and Algorithms

The minimum height of a binary tree is determined as follows:

min 2log 1H N

For instance, if there are three nodes to be stored in the binary tree (N=3) then Hmin=2.

15Data Structures and Algorithms

Given a height of the binary tree, H, the minimum number of nodes in the tree is given as follows:

minN H

16Data Structures and Algorithms

Given a height of the binary tree, H, the maximum number of nodes in the tree is given as follows:

max 2 1HN

17Data Structures and Algorithms

A complete tree has the maximum number of entries for its height. The maximum number is reached when the last level is full.

A tree is considered nearly complete if it has the minimum height for its nodes and all nodes in the last level are found on the left

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Sequential Representation

AB--C------D--.E

[1][2][3][4][5][6][7][8][9].[16]

[1][2][3][4][5][6][7][8][9]

ABCDEFGHI

A

B

E

C

D

A

B C

GE

I

D

H

F

(1) waste space(2) insertion/deletion problem

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Linked Representationtypedef struct tnode *ptnode;

typedef struct tnode {

int data;

ptnode left, right;

};

dataleft right

data

left right

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04/12/23 04:50 AM 23Data Structures and Algorithms

A binary tree traversal requires that each node of the tree be processed once and only once in a predetermined sequence.

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Preorder Traversal (recursive version)

void preorder(tnode ptr)

/* preorder tree traversal */

{

if (ptr) {

printf(“%d”, ptr->data);

preorder(ptr->left);

predorder(ptr->right);

}

}

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Inorder Traversal (recursive version)

void inorder(tnode ptr)

/* inorder tree traversal */

{

if (ptr) {

inorder(ptr->left);

printf(“%d”, ptr->data);

indorder(ptr->right);

}

}

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Postorder Traversal (recursive version)

void postorder(tnode ptr)

/* postorder tree traversal */

{

if (ptr) {

postorder(ptr->left);

postdorder(ptr->right);

printf(“%d”, ptr->data);

}

}

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Using Pre order and Post order Traversal

Using Post order and In order Traversal

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1. Scan the preorder traversal from left to right.

2. For each scanned node locate its position in inorder traversal. Let the scanned node be X.

3. The nodes preceding X in inorder form its left subtree and nodes succeeding it form right subtree.

4. Repeat step1 for each symbol in the preorder.

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Preorder: A B D H E C F G

Inorder: D H B E A F C G

A

B C

D E

H

F G

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1. Scan the post order traversal from right to left.

2. For each node scanned locate its position in order traversal. Let the scanned node be X.

3. The node preceding X in inorder, form its left subtree and nodes succeeding it form right subtree.

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A

B C

D E

I

F G

Postorder: H D I E B J F K L G C A

Inorder: H D B I E A F J C K G L

H JK L

04/12/23 04:50 AM 41Data Structures and Algorithms

A variable is an expression. If x and y are expressions, then ¬x, xy,

xy are expressions. Parentheses can be used to alter the

normal order of evaluation (¬ > > ). Example: x1 (x2 ¬x3)

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X3X1

X2 X1

X3

(x1 ¬x2) (¬ x1 x3) ¬x3

postorder traversal (postfix evaluation)04/12/23 04:50 AM 43Data Structures and Algorithms

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