Tree Data Structure

1. Unit – VI Tree
Prepared By:
Dabbal Singh Mahara
1

2. a. Concept and definition
b. Binary tree
c. Introduction and application
d. Operations
e. 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 Tree
i. Huffman algorithm
j. AVL Trees
Tree data structure
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3. 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.

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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5. 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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6. • 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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7. 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.
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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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9.  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.
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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
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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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12. B
J
G
F
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).
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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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14. • Array representation
• Linked Representation
Binary Tree Representation
Array Representation
An 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
G
F
C
I
ED
H
A0
1 2
4
7
5
14
6
3
10
Declare array of size 2 3+1 -1 = 15
Char a[15]
A B C D E F G H I J
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 14By Dabal Mahara

15. 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 tree
struct binaryTreeNode
{
int info;
struct binaryTreenode *left;
struct binaryTreenode *right;
};
struct binaryTreenode *root=NULL;
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16. The given tree can be represented in linked list as
given below.
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17.  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 traversal
ii. In-order traversal
iii. Post-order traversal
Tree traversal
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18. 1. Pre-order traversal:
The preorder traversal of a nonempty binary tree is defined as
follows:
i. Visit the root node
ii. Traverse the left sub-tree in preorder
iii. Traverse the right sub-tree in preorder
The preorder traversal output of the given tree
is: A B D H I E C F G
The 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);
}
}
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19. The inorder traversal of a nonempty binary tree is defined as follows:
i. Traverse the left sub-tree in inorder
ii. Visit the root node
iii. 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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20. 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-order
ii. Traverse the right sub-tree in post-order
iii. 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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21. Q. Design a binary tree whose traversal sequences are
given below:
Pre-order: CPRBLOEMATIK
In-Order: BRLPEOMCTAIK
Solution:
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
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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.
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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
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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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25. 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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26. S
P T
MV Q
R
Step -4: The final tree is given below.
O
K
B
A
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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27. 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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28. 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.
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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.
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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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31. struct bstnode
{
int info; //Declaring an info field
struct bstnode *left; // left child pointer
struct bstnode *right; // right child pointer
};
Structure of BST node
Struct bstnode *root = NULL;
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32. 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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33. 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);
}
33By Dabal Mahara

34. 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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35. 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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36. 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);
}
}
36By Dabal Mahara

37. 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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38. 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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39. 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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40. 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 or
b. left most node of it's right sub-tree.
6. End
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41. 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);
} 41By Dabal Mahara

42. struct bstnode * findmin(struct bstnode * root)
{
if(root==NULL)
return NULL;
else if(root->left ==NULL)
return head;
else
return(findmin(root->left));
}
42By Dabal Mahara

43. Expression Tree
An 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)
43By Dabal Mahara

44. Expression Tree Examples
Inorder Traversal ResultExpression TreeExpression
a + 3(a+3)
3+4*5-9+63+(4*5-(9+6))
log xlog(x)
n !n!
+
3a
+
-3
*
54
+
69
log
x
!
n
44By Dabal Mahara

45. 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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46. 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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47. 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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48. 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+$
48By Dabal Mahara

49. • 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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50. 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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51. Example: 1
Let 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 frequency
A 3
O 5
T 7
E 10
51By Dabal Mahara

52. 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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53. 4. Repeat the step 2 again.
T3: 25
E: 10
T2 : 15
T : 7
T1 : 8
O: 5A : 3
T3: 25
E: 10
T2 : 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.
53By Dabal Mahara

54. Now the code for each characters are as obtained by collecting the bits from
root to the character node. The codes are:
Char code frequency
E 0 10
T 10 7
A 110 3
O 111 5
Therefore, total number of bits required = 10* 1 +7 *2 + 3 *3 + 5 *3 =48 bits
The space saved = (50-48)/50 *100 % = 4 %
54By Dabal Mahara

55. Example: 2
Let us take any four characters and their frequencies as follows:
Char frequency
a 10
e 15
i 12
s 3
t 4
Sp (space) 13
nl (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 frequency
nl (new line) 1
s 3
t 4
a 10
i 12
Sp (space) 13
e 15
55By Dabal Mahara

56. 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 : 3
nl :
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:
13
e:
15
T2 : 8
T1 : 4
S : 3
nl :
1
t :
4
56By Dabal Mahara

57. 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
57By Dabal Mahara

58. 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
58By Dabal Mahara

59. 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.
59By Dabal Mahara

60. 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.
0
1
0
0
0
0
0 1
1
1
1
1
This tree is called Huffman tree.
60By Dabal Mahara

61. Now the code for each characters are as obtained by collecting the bits from
root to the character node. The codes are:
Char code frequency
i 00 12
sp 01 13
e 10 15
nl 11000 1
S 11001 3
t 1101 4
a 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 %
61By Dabal Mahara

62. 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.
88
44
17 78
32 50
48 62
2
4
1
1
2
3
1
1
An example of an AVL tree where the
heights are shown next to the nodes:
62By Dabal Mahara

63. 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)
63By Dabal Mahara

64. 64By Dabal Mahara