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Unit – VI Tree
Prepared By:Dabbal Singh Mahara
By Dabal Mahara 2
a. Concept and definitionb. Binary tree c. Introduction and application d. Operationse. Types of binary tree
Complete binary tree Strictly binary tree Almost complete binary tree
f. Tree traversal Pre-order traversal In-order traversal Post-order traversal
g. Binary search tree searching insertion deletion
h. Expression Treei. Huffman algorithm j. AVL Trees
Tree data structure
Introduction
• A tree is an abstract model of a hierarchical structure that consists of nodes with a parent-child relationship. A tree is a finite set of data items or nodes such that: There is a special data-item called the root of the tree Its remaining data items are partitioned into mutually exclusive sub-
sets, each of which itself is a tree, called sub-trees.
By Dabal Mahara 4
Characteristics of trees: Non-linear data structure Combines advantages of an ordered array Searching as fast as in ordered array Insertion and deletion as fast as in linked list
Application: Directory structure of a file store Structure of arithmetic expressions Hierarchy of an organization
Some key terms Degree of a node: The degree of a node is the number of children of that node. In above tree ,
degree(A) = 2. Degree of a Tree: The degree of a tree is the maximum degree of nodes in a given tree. In the
above tree, maximum degree of nodes is 2, thus the degree of the tree is 2. Path: It is the sequence of consecutive edges from source node to destination node.
There is a single unique path from the root to any node.
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Height of a node:The height of a node is the maximum path length from that node to a leaf node. A leaf node has a height of 0.
Height of a tree:The height of a tree is the height of the root.
Depth of a node: Depth of a node is the path length from the root to that node. The
root node has a depth of 0.
Depth of a tree:Depth of a tree is the maximum level of any leaf in the tree. This is equal to the longest path from the root to any leaf.
Level of a node:The level of a node is 0, if it is root; otherwise it is one more then
its parent.
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• A is the root node• B is the parent of E and F• D is the sibling of B and C• E and F are children of B• E, F, G, D are external nodes or leaves• A, B, C are internal nodes• Depth of F is 2• the height of tree is 2• the degree of node A is 3• The degree of tree is 3
Illustration:
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Binary Trees A binary tree is a finite set of elements that are either empty or is
partitioned into three disjoint subsets. The first subset contains a single element called the root of the tree. The other two subsets are themselves binary trees called the left
and right sub-trees of the original tree. A left or right sub tree can be empty.
Each element of a binary tree is called a node of the tree. The following figure shows a binary tree with 9 nodes where A is the root.
By Dabal Mahara 8
Binary tree properties: If a binary tree contains m nodes at level l , it contains at
most 2m nodes at level l + 1. Since a binary tree can contain at most 1 node at level 0 (the
root), it contains at most 2l nodes at level l.
Types of binary tree
Strictly binary tree
Almost complete binary tree
Complete binary tree
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If every non-leaf node in a binary tree has non-empty left and right sub-trees, then such a tree is called a strictly binary tree.
Strictly binary tree
The following tree is not strictly binary tree, Since node E has a left son but not a right son.
By Dabal Mahara 10
A strictly binary tree of depth d is called complete binary tree if all of whose leaves are at level d.
A complete binary tree with depth d has 2d leaves and 2d -1 non-leaf nodes (internal).
Complete binary tree:
Example: In this tree d = 3, therefore, leaves = 23 = 8 Internal nodes = 23 -1 = 7
By Dabal Mahara 11
A binary tree of depth d is an almost complete binary tree if:i. Any node at level less than d-1 has two sons.ii. For any node in the tree with a right descendant at level d, the node must
have a left son and every left descendant of the node is either a leaf at level d or has two sons.
Almost complete binary tree:
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B
J
GF
C
I
ED
H
A
K
Level 0
Level 1
Level 2
Level 3
This tree is not almost complete binary tree because it does not satisfy condition ii. That is, node A has right descendent at level 3 ( J or K), but it has left descendent leaf node at level 2 (node E).
By Dabal Mahara 13
Operations on Binary tree: father(n,T): Return the parent node of the node n in tree T. If n is the
root, NULL is returned. LeftChild(n,T): Return the left child of node n in tree T. Return NULL if n
does not have a left child. RightChild(n,T): Return the right child of node n in tree T. Return NULL if n
does not have a right child. Info(n,T): Return information stored in node n of tree T (ie. Content of
a node). Sibling(n,T): return the sibling node of node n in tree T. Return NULL if n
has no sibling. Root(T): Return root node of a tree if and only if the tree is
nonempty. Size(T): Return the number of nodes in tree T MakeEmpty(T): Create an empty tree T SetLeft(S,T): Attach the tree S as the left sub-tree of tree T SetRight(S,T): Attach the tree S as the right sub-tree of tree T. Preorder(T): Traverses all the nodes of tree T in preorder. postorder(T): Traverses all the nodes of tree T in postorder Inorder(T): Traverses all the nodes of tree T in inorder.
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• Array representation• Linked Representation
Binary Tree Representation
Array RepresentationAn array is used to store the nodes of the binary tree. Nodes stored in the array are accessed sequentially, root is at index 0 and then left child and right child are stored. To store the elements of the binary tree of depth d, declare the array of size 2d+1 -1, i.e. the total number of nodes of complete binary tree and store the elements level wise. If root is at I then left child is t 2i+1 and right child is at 2i+2.
B
J
GF
C
I
ED
H
A0
1 2
4
7
5
14
63
10
Declare array of size 2 3+1 -1 = 15 Char a[15]
A B C D E F G H I J0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
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Linked Representation of Binary Tree• It is the most popular way to present a binary tree. Each element is
represented by a node that has two link fields (leftChild and rightChild) plus an element field. Each binary tree node is represented as an object whose data type is binaryTreeNode .
• The space required by an n node binary tree is: n * sizeof(binaryTreeNode)
C representation for Binary treestruct binaryTreeNode{
int info;struct binaryTreenode *left;struct binaryTreenode *right;
};
struct binaryTreenode *root=NULL;
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The given tree can be represented in linked list as given below.
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The tree traversal is a way in which each node in the tree is visited exactly once in a symmetric manner.
There are three popular methods of traversal
i. Pre-order traversalii. In-order traversaliii. Post-order traversal
Tree traversal
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1. Pre-order traversal:The preorder traversal of a nonempty binary tree is defined as follows:i. Visit the root nodeii. Traverse the left sub-tree in preorderiii. Traverse the right sub-tree in preorder
The preorder traversal output of the given tree is: A B D H I E C F GThe preorder is also known as depth first order.
C function for preorder traversing: void preorder(struct bnode *root) {
if(root!=NULL) {
printf(“%c”, root->info);preorder(root->left);preorder(root->right);
} }
By Dabal Mahara 19
The inorder traversal of a nonempty binary tree is defined as follows:i. Traverse the left sub-tree in inorderii. Visit the root nodeiii. Traverse the right sub-tree in inorder
The inorder traversal output of the given tree is: H D I B E A F C G
C function for inorder traversing: void inorder(struct binaryTreenode *root) {
if(root!=NULL) {
inorder(root->left);printf(“%c”, root->info); inorder(root->right);
} }
2. In-order traversal:
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3. Post-order traversal:The post-order traversal of a nonempty binary tree is defined as follows:
i. Traverse the left sub-tree in post-orderii. Traverse the right sub-tree in post-orderiii. Visit the root node
C function for post-order traversing: void post-order(struct bnode *root) {
if(root!=NULL) {
post-order(root->left);post-order(root->right);printf(“%c”, root->info);
} }
The post-order traversal output of the given tree is: H I D E B F G C A
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Q. Design a binary tree whose traversal sequences are given below:
Pre-order: CPRBLOEMATIKIn-Order: BRLPEOMCTAIKSolution:
Since preorder traverses from root of the tree, C is root of the tree. The { B, R, L, P, E, O, M} and {T, I, A, K } are respectively left and right sub tree . (from inorder sequence)
C
T, A, I, K
B, R, L, P, E, O,
M
By Dabal Mahara 22
C
I, KB, R, L,
P
E, O, M
A
T
Similarly, for left sub tree { B, R, L, P, E, O, M}, root is P ( from preorder sequence) and {B, R, L} is left sub-tree and {E, O, M} is right sub-tree. ( From inorder sequence) For right sub-tree {T, A, I, K}, A is the root. ( from preorder sequence) T is left sub-tree and {I, K } is right sub-tree.
By Dabal Mahara 23
Similarly, R is root for sub-tree {B, R, L} ( from preorder sequence)B is left child and L right child of R ( from inorder sequence)O is root for sub-tree {E, O, M} (from preorder sequence)E is left child and M is right child of O (from inorder sequence)In the same way, I is root and K is right child for sub-tree { I, K }
C
P A
TO I
KL M
R
EB
By Dabal Mahara 24
Q. A binary tree T has 12 nodes. The in-order and post-order traversals of T yield the following sequence of nodes:
In-order : VPNAQRSOKBTM Post-order : VANRQPBKOMTS Construct the Binary tree T showing each step. Explain how you can
arrive at solution in brief.
Solution:Since post-order traversal visits all nodes in left sub-tree , right sub-tree and root recursively, the last node in the sequence is root of the tree. That is, S is the root.In-order traversal visits the nodes in Left child, root and right child in recursive way, { V,P, N,A, Q, R} is left sub-tree and {O, K, B, T, M} is the right sub tree.
S
O, K, B, T, M
V, P, N, A, Q, R
Step -1:
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S
O, K, B
P
T
MV N, A, Q, R
O
S
K, B
P
N, A
T
MV Q
R
Step -2: The root of { V, P, N, A, Q, R} is P obtained from postorder sequence. Left subtree and right sub trees are {v} and {N, A, Q, R} obtained from inorder sequence. Similarly, for { O, K, T, B, M} , root is T and left sub tree {O, K, B} and right sub tree is {M}.
Step -3: In the same way, root and sub trees are obtained.
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S
P T
MV Q
R
Step -4: The final tree is given below.
O
K
BA
N
Thus, root of the tree is obtained from post-order sequence scanning from right to left and left and right sub-trees of corresponding root are obtained from inorder sequence.
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An important application of binary trees is their use in searching.A binary search tree (BST) is a binary tree that is either empty or in which every node contains a key (value) and satisfies the following conditions:
i. All keys in the left sub-tree of the root are smaller than the key in the root node.
ii. All keys in the right sub-tree of the root are greater than the key in the root node.
iii. The left and right sub-trees of the root are again binary search trees .
Binary search tree(BST)
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Binary Search Trees
1 4
2
3
8
6
1 4
2
3
8
6
7
• The tree on the left is a binary search tree, but the tree on the right is not.
By Dabal Mahara 29
Given the following sequence of numbers, 14, 15, 4, 9, 7, 18, 3, 5, 16, 20, 17, 5 The following binary search tree can be constructed:
14
4
9
15
7
5 17
2016
3 18
The inorder traversal of BST gives the sorted sequence in ascending order. So, inorder traversal of above BST is: 3, 4, 5, 7, 9, 14, 15, 16, 17, 18, 20.
By Dabal Mahara 30
Following operations can be done in BST: Search(k, T): Search for key k in the tree T. If k is found in some node
of tree then return true otherwise return false.
Insert(k, T): Insert a new node with value k in the info field in the tree T such that the property of BST is maintained.
Delete(k, T):Delete a node with value k in the info field from the tree T such that the property of BST is maintained.
FindMin(T), FindMax(T): Find minimum and maximum element from the given nonempty BST.
Operations on Binary search tree(BST):
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struct bstnode{
int info; //Declaring an info fieldstruct bstnode *left; // left child pointer struct bstnode *right; // right child pointer
};
Structure of BST node
Struct bstnode *root = NULL;
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To find if any node of a BST contains an element equal to target: 1. Set curr to the BST’s root. 2. Repeat:
2.1. If curr is null: 2.1.1. Terminate with answer none.
2.2. Otherwise, if target is equal to curr’s element: 2.2.1. Terminate with answer curr.
2.3. Otherwise, if target is less than curr’s element: 2.3.1. Set curr to curr’s left child.
2.4. Otherwise, if target is greater than curr’s element: 2.4.1. Set curr to curr’s right child.
2. end
BST search algorithm :
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C function for BST searching: void BinSearch(struct bstnode *root , int key) {
if(root == NULL) {
printf(“The number does not exist”); } else if (key == root->info) {
printf(“The searched item is found”): } else if(key < root->info)
BinSearch(root->left, key); else
BinSearch(root->right, key); }
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To insert a new item in a tree, we must first verify that its key is different from those of existing elements. To do this a search is carried out. If the search is unsuccessful, then item is inserted.
Idea: To insert a new element into a BST, proceed as if searching for that element. If the element is not already present, the search will lead to a null link. Replace that null link by a link to a leaf node containing the new element.
Insertion of a node in BST:
15
9
13
12
10
18
20
22
25
Insertion of element 13
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To insert the element elem into a BST: 1. Set parent to null, and set curr to the BST’s root.2. Repeat:
2.1. If curr is null:2.1.1. Replace the null link from which curr was taken (either the BST’s root or parent’s left child or parent’s right child) link to a newly-created leaf node with element elem.2.1.2. Terminate.
2.2. Otherwise, if elem is equal to curr’s element:2.2.1. Terminate.
2.3. Otherwise, if elem is less than curr’s element:2.3.1. Set parent to curr, and set curr to curr’s left child.
2.4. Otherwise, if elem is greater than curr’s element:2.4.1. Set parent to curr, and set curr to curr’s right child.
3.End
BST insertion algorithm :
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C function for BST insertion:Struct bstnode * insert(struct bstnode *root, int item){
if(root=NULL){
root=(struct bstnode*)malloc (sizeof(struct bnode));root->left=root->right=NULL;root->info=item;
}else if ( item == root->info)
{ printf("\n Duplicate not allowed."); return root; }
else{
if(item<root->info) root->left=insert(root->left, item);else
root->right=insert(root->right, item);}
}
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While deleting a node from BST, there may be three cases:
1. The node to be deleted may be a leaf node:In this case simply delete a node and set null pointer to its parents those side at which this deleted node exist.
Deleting a node from the BST:
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2. The node to be deleted has one child: In this case the child of the node to be deleted is appended to its parent
node. Suppose node to be deleted is 18
Delete(18)
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3. The node to be deleted has two children:In this case node to be deleted is replaced by its in-order successor node. That is, the node to be deleted is either replaced by its right sub-trees leftmost node or its left sub-trees rightmost node.Suppose node to deleted is 12. Find minimum element in the right sub-tree of the node to be removed. In current example it is 19.
Delete(12)
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General algorithm to delete a node from a BST:
1. start 2. if a node to be deleted is a leaf node at left side then simply delete and set null
pointer to it's parent's left pointer.3. If a node to be deleted is a leaf node at right side then simply delete and set
null pointer to it's parent's right pointer.4. if a node to be deleted has one child then connect it's child pointer with it's
parent pointer and delete it from the tree.5. if a node to be deleted has two children then replace the node being deleted
either by a. right most node of it's left sub-tree orb. left most node of it's right sub-tree.
6. End
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The delnode function:
struct bstnode *delnode(struct bstnode *root, int item) {
struct bstnode *temp;if(root==NULL){
printf(“Empty tree”);return;
}else if(item<root->info)
root->left=delnode(root->left, item);else if(item>root->info)
root->right=delnode(root->right, item);else if(root->left!=NULL && root->right!=NULL) //node has two children{
temp=findmin(root->right);root->info=temp->info;root->right=delnode(root->right, root->info);
}else{
temp=root;if(root->left==NULL)
root=root->right;else if(root->right==NULL)
root=root->left;free(temp);
}return(root);
}
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struct bstnode * findmin(struct bstnode * root){
if(root==NULL) return NULL;else if(root->left ==NULL) return head;else return(findmin(root->left));
}
By Dabal Mahara 43
Expression TreeAn expression tree is a strictly binary tree in which leaf node contains operands and non-leaf node contain operators. Root nodes contain the operators that is applied to the result of left and right sub trees. The operators, constants, and variables are arranged in such a way that an inorder traversal of the tree produces the original expression without parentheses.
Example: 1 An expression tree of an expression (a + b *c ) +((c*e +f) *g)
*a
+ *
g+f*
ec
c
+
b
Exercise: Construct an expression tree of an expression: (a + b ) /((c-d) *e)
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Expression Tree ExamplesInorder Traversal Result Expression Tree Expression
a + 3 (a+3)
3+4*5-9+6 3+(4*5-(9+6))
log x log(x)
n ! n!
+
3a
+
-3
*54
+
69
log
x
!
n
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Constructing an Expression Tree From Postfix Expression Algorithm • We read the expression one symbol at a time.
• If the symbol is an operand, we create a one-node tree and push a pointer to it
onto a stack.
• If the symbol is an operator, we pop pointers to two trees T1 and T2 from the stack
(T1 is popped first) and form a new tree whose root is the operator and whose left
and right children point to T2 and T1, respectively. A pointer to this new tree is then
pushed onto the stack.
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Example: • Input: ab+cde+**
The first two symbols are operands, so we create one-node trees and push pointers to them onto a stack. For convenience, we will have the stack grow from left to right in the diagrams.
a b
Next, a ‘+’ is read, so two pointers to trees are popped, a new tree is formed, and a pointer to it is pushed onto the stack.
a b
+
Next, c, d, and e are read, and for each a one-node tree is created and a pointer to the corresponding tree is pushed onto the stack.
a b
+ c d e
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Constructing an Expression Tree
a b
+ c
• Now a ‘+’ is read, so two trees are merged.
d e
+
a b
+
c
Continuing, a ‘*’ is read, so we pop two tree pointers and form a new tree with a ‘*’ as root.
d e
+
*
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Constructing an Expression Tree
a b
+
c
Finally, the last symbol is read, two trees are merged, and a pointer to the final tree is left on the stack.
d e
+
*
*
Exercise: Construct an expression from postfix expression: AB+C*DC--FG+$
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• Proposed by Dr. David A. Huffman at MIT in 1952 – “A Method for the Construction of Minimum Redundancy Codes”
• Huffman coding is a form of statistical coding– more frequently used symbols have shorter code words– Not all characters occur with the same frequency! – Yet all characters are allocated the same amount of space – 1 char = 1 byte, be it e or x
• Huffman coding is a technique used to compress files for transmission• Works well for text and fax transmissions
The Huffman Coding
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Huffman Algorithm 1. Initially, each symbol is one node tree by itself. That is, there is forest of n nodes
each labelled with symbol and weight.2. Repeat while n>1
i. Select the two trees T1 and T2 with minimum weights from the forest.ii. Merge T1 and T2 to form T3 with weight sum of both trees.iii. Insert T3 into the forest.
3. Assign 0 to each left path of tree and 1 to right path. The code for each symbol is obtained by collecting bits on the paths from root to the leaf node labelled with the symbol.
• Huffman codes is text compression method, which relies on the relative frequency (i.e., the number of occurrences of a symbol) with which different symbols appear in a text
• Uses extended binary trees • Variable-length codes that satisfy the property, where no code is a prefix of another• Huffman tree is a binary tree with minimum weighted external path length for a
given set of frequencies (weights).
Huffman Code
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Example: 1Let us take any four characters and their frequencies as follows:Char frequency E 10 T 7 O 5 A 3
• Without using Huffman coding, these four characters can be represented by using two bits as: 00 -> E, T ->01, O ->10 and A-> 11. This is fixed length coding. So, bits required to encode all the characters in the above message will be:
( 10 +7+5+3) * 2 = 50 bits • Huffman coding technique uses variable length code depending upon the frequency of characters used in the text. It's coding technique is as follows:
First sort the characters by their frequency in non-decreasing order.Char frequencyA 3O 5T 7E 10
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1. Create a forest of one node trees for each character as:
A : 3 O : 5 T : 7 E: 10
2. Merge two minimum trees as a single tree and add it to the forest in the order.
T1 : 8
O: 5A : 3
T : 7 E: 10
3. Repeat step 2 again.
T2 : 15
T : 7
T1 : 8
O: 5A : 3
E: 10
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4. Repeat the step 2 again.
T3: 25
E: 10T2 : 15
T : 7T1 : 8
O: 5A : 3
T3: 25
E: 10T2 : 15
T : 7
T1 : 8
O: 5A : 3
5. Assign 0 to left path and 1 to right path in the tree.
0
0
0
1
1
1
This tree is called Huffman tree.
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Now the code for each characters are as obtained by collecting the bits from root to the character node. The codes are:Char code frequencyE 0 10T 10 7A 110 3O 111 5
Therefore, total number of bits required = 10* 1 +7 *2 + 3 *3 + 5 *3 =48 bitsThe space saved = (50-48)/50 *100 % = 4 %
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Example: 2Let us take any four characters and their frequencies as follows:Char frequency a 10 e 15 i 12s 3t 4Sp (space) 13nl (new line) 1
Fixed Length Format: These 7 characters can be represented by three bits.So, total number of bits = (10+15+12+3+4+13+1) * 3 = 174
Using Huffman Coding First sort the characters by their frequency in non-decreasing order.Char frequencynl (new line) 1s 3t 4a 10i 12Sp (space) 13 e 15
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1. Create a forest of one node trees for each character as:
2. Merge two minimum trees as a single tree and add it to the forest in the order.
T1 : 4
S : 3nl : 1
3. Repeat step 2 again.
nl : 1
s: 3
t : 4
a: 10
i: 12
sp: 13
e: 15
t : 4
a: 10
i: 12
sp: 13
e: 15
a: 10 i: 12 sp:
13e: 15
T2 : 8
T1 : 4
S : 3nl : 1
t : 4
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4. Repeat step 2 again.
i: 12 sp: 13 e: 15
T2 : 8
T1 : 4
S : 3nl : 1
t : 4
T3 : 18
a : 10
5. Repeat step 2 again.
e: 15 T4: 25
sp: 13i: 12
T2 : 8
T1 : 4
S : 3nl : 1
t : 4
T3 : 18
a : 10
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6. Repeat step 2 again.
sp: 13
T4: 25
i: 12 T2 : 8
T1 : 4
S : 3nl : 1
t : 4
T3 : 18
a : 10
e: 15
T5: 33
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sp: 13
T4: 25
i: 12 T2 : 8
T1 : 4
S : 3nl : 1
t : 4
T3 : 18
a : 10
e: 15
T5: 33
T6: 58
7. Repeat step 2 again.
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sp: 13
T4: 25
i: 12 T2 : 8
T1 : 4
S : 3nl : 1
t : 4
T3 : 18
a : 10
e: 15
T5: 33
T6: 58
8. Assign 0 to left path and 1 to right path in the tree.
01
0
00
0
0 1
1
1
1
1
This tree is called Huffman tree.
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Now the code for each characters are as obtained by collecting the bits from root to the character node. The codes are:Char code frequencyi 00 12sp 01 13e 10 15nl 11000 1S 11001 3t 1101 4a 111 10
Therefore, total number of bits required = 12*2 +13 *2 + 15 *2 + 1 *5 +3 * 5 + 4*4 + 10 * 3 = 146 bits
The space saved = (174-146)/174 *100 % = 16.09 %
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AVL (Adelson-Velskii and Landis) Trees
• An AVL Tree is a binary search tree such that for every internal node v of T, the heights of the children of v can differ by at most 1.
An example of an AVL tree where the heights are shown next to the nodes:
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5
3 8
1 4 10
AVL Tree
5
3
1 4
Not AVL Tree
• All operations depend on the depth of the tree.• BSTs are limited because of their bad worst-case performance O(n).• AVL trees are height-balanced binary search trees• An AVL tree has balance factor calculated at every node
For every node, heights of left and right subtree can differ by no more than 1
• It allows dynamic insert and remove with O(log(N)) time complexity.• Balanced search trees are trees whose heights in the worst case is O(lg n)
64
Thank You !
By Dabal Mahara
