1.
Presented By:
Susheel Thakur
M.Tech(C.S)
Roll No- 324
2.
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.
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Data Structures and Algorithms
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3.
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
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Tree
Data Structures and Algorithms
Binary Tree
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4.
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
Interior nodes
Height
Leaf nodes
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5.
General Tree – organization chart format
Parenthetical Listing
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6.
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7.
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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8.
A (B (C D) E F (G H I) )
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9.
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
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link 1
link 2
...
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link n
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10.
Every tree node:
object – useful information
children – pointers to its children nodes
O
O
O
O
O
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11.
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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12.
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13.
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
H max = N
A tree with a maximum height is rare. It occurs when all of the nodes in the
entire tree have only one successor.
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15.
The minimum height of a binary tree is
determined as follows:
H min = [ log 2 N ] + 1
For instance, if there are three nodes to be stored in the binary tree (N=3)
then Hmin=2.
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16.
Given a height of the binary tree, H, the
minimum number of nodes in the tree is
given as follows:
N min = H
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17.
Given a height of the binary tree, H, the
maximum number of nodes in the tree is
given as follows:
N max = 2 − 1
H
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18.
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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19.
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20.
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21.
Sequential
Representation
A
B
C
D
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
.
[16]
(1) waste space
(2) insertion/deletion
problem
A
B
-C
---D
-.
E
A
B
C
D
E
F
G
H
I
A
B
D
E
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[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
H
I
Data Structures and Algorithms
C
E
F
G
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22.
Linked Representation
typedef struct tnode *ptnode;
typedef struct tnode {
int data;
ptnode left, right;
};
data
left
data
right
left
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right
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23.
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24.
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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25.
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26.
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27.
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28.
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29.
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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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31.
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32.
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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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34.
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35.
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36.
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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37.
Using Pre order and Post order Traversal
Using Post order and In order Traversal
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38.
1.
2.
3.
4.
Scan the preorder traversal from left to right.
For each scanned node locate its position in
inorder traversal. Let the scanned node be X.
The nodes preceding X in inorder form its left
subtree and nodes succeeding it form right
subtree.
Repeat step1 for each symbol in the preorder.
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39.
Preorder:
ABDHECFG
Inorder:
DHBEAFCG
A
B
D
C
E
F
G
H
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40.
1.
2.
3.
Scan the post order traversal from right to left.
For each node scanned locate its position in
order traversal. Let the scanned node be X.
The node preceding X in inorder, form its left
subtree and nodes succeeding it form right
subtree.
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41.
Postorder:
HDIEBJFKLGCA
Inorder:
HDBIEAFJCKGL
A
B
D
H
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C
E
I
F
G
J
Data Structures and Algorithms
K
L
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42.
A variable is an expression.
If x and y are expressions, then ¬x, x∧y,
x∨y are expressions.
Parentheses can be used to alter the normal
order of evaluation (¬ > ∧ > ∨).
Example: x1 ∨ (x2 ∧ ¬x3)
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