Tag: AVL Tree Rotation Examples PPT

AVL Tree Properties | Problems on AVL Tree

AVL Tree-

 

Before you go through this article, make sure that you have gone through the previous article on AVL Trees.

 

We have discussed-

  • AVL trees are self-balancing binary search trees.
  • In AVL trees, balancing factor of each node is either 0 or 1 or -1.

 

In this article, we will discuss AVL Tree Properties.

 

AVL Tree Properties-

 

Important properties of AVL tree are-

 

Property-01:

 

Maximum possible number of nodes in AVL tree of height H

= 2H+1 – 1

 

Example-

 

Maximum possible number of nodes in AVL tree of height-3

= 23+1 – 1

= 16 – 1

= 15

Thus, in AVL tree of height-3, maximum number of nodes that can be inserted = 15.

 

 

We can not insert more number of nodes in this AVL tree.

 

Property-02:

 

Minimum number of nodes in AVL Tree of height H is given by a recursive relation-

N(H) = N(H-1) + N(H-2) + 1

 

Base conditions for this recursive relation are-

  • N(0) = 1
  • N(1) = 2

 

Example-

 

Minimum possible number of nodes in AVL tree of height-3 = 7

 

 

(For explanation, Refer problem-01)

 

Property-03:

 

Minimum possible height of AVL Tree using N nodes

= ⌊log2N⌋

 

Example-

 

Minimum possible height of AVL Tree using 8 nodes

= ⌊log28⌋

= ⌊log223

= ⌊3log22⌋

= ⌊3⌋

= 3

 

Property-04:

 

Maximum height of AVL Tree using N nodes is calculated using recursive relation-

N(H) = N(H-1) + N(H-2) + 1

 

Base conditions for this recursive relation are-

  • N(0) = 1
  • N(1) = 2

 

NOTE-

 

  • If there are n nodes in AVL Tree, its maximum height can not exceed 1.44log2n.
  • In other words, Worst case height of AVL Tree with n nodes = 1.44log2n.

 

PRACTICE PROBLEMS BASED ON AVL TREE PROPERTIES-

 

Problem-01:

 

Find the minimum number of nodes required to construct AVL Tree of height = 3.

 

Solution-

 

We know, minimum number of nodes in AVL tree of height H is given by a recursive relation-

 

N(H) = N(H-1) + N(H-2) + 1

 

where N(0) = 1 and N(1) = 2

 

Step-01:

 

Substituting H = 3 in the recursive relation, we get-

N(3) = N(3-1) + N(3-2) + 1

N(3) = N(2) + N(1) + 1

N(3) = N(2) + 2 + 1                (Using base condition)

N(3) = N(2) + 3                      …………(1)

 

To solve this recursive relation, we need the value of N(2).

 

Step-02:

 

Substituting H = 2 in the recursive relation, we get-

N(2) = N(2-1) + N(2-2) + 1

N(2) = N(1) + N(0) + 1

N(2) = 2 + 1 + 1                 (Using base conditions)

∴ N(2) = 4                          …………(2)

 

Step-03:

 

Using (2) in (1), we get-

N(3) = 4 + 3

∴ N(3) = 7

Thus, minimum number of nodes required to construct AVL tree of height-3 = 7.

 

Problem-02:

 

Find the minimum number of nodes required to construct AVL Tree of height = 4.

 

Solution-

 

We know, minimum number of nodes in AVL tree of height H is given by a recursive relation-

 

N(H) = N(H-1) + N(H-2) + 1

 

where N(0) = 1 and N(1) = 2

 

Step-01:

 

Substituting H = 4 in the recursive relation, we get-

N(4) = N(4-1) + N(4-2) + 1

N(4) = N(3) + N(2) + 1               …………(1)

 

To solve this recursive relation, we need the value of N(2) and N(3).

 

Step-02:

 

Substituting H = 2 in the recursive relation, we get-

N(2) = N(2-1) + N(2-2) + 1

N(2) = N(1) + N(0) + 1

N(2) = 2 + 1 + 1                 (Using base conditions)

∴ N(2) = 4                          …………(2)

 

Step-03:

 

Substituting H = 3 in the recursive relation, we get-

N(3) = N(3-1) + N(3-2) + 1

N(3) = N(2) + N(1) + 1

N(3) = 4 + 2 + 1                 (Using (2) and base condition)

∴ N(3) = 7                          …………(3)

 

Step-04:

 

Using (2) and (3) in (1), we get-

N(4) = 7 + 4 + 1

∴ N(4) = 12

Thus, minimum number of nodes required to construct AVL tree of height-4 = 12.

 

Problem-03:

 

What is the maximum height of any AVL tree with 10 nodes?

 

Solution-

 

For calculating the maximum height of AVL tree with n nodes, we use a recursive relation-

 

N(H) = N(H-1) + N(H-2) + 1

 

Step-01:

 

Substituting H = 2 in the recursive relation, we get-

N(2) = N(2-1) + N(2-2) + 1

N(2) = N(1) + N(0) + 1

N(2) = 2 + 1 + 1                (Using base conditions)

∴ N(2) = 4                          …………(1)

 

So, minimum number of nodes required to construct AVL tree of height-2 = 4.

 

Step-02:

 

Substituting H = 3 in the recursive relation, we get-

N(3) = N(3-1) + N(3-2) + 1

N(3) = N(2) + N(1) + 1

N(3) = 4 + 2 + 1                (Using (1) and base condition)

∴ N(3) = 7                          …………(2)

 

So, minimum number of nodes required to construct AVL tree of height-3 = 7.

 

Step-03:

 

Substituting H = 4 in the recursive relation, we get-

N(4) = N(4-1) + N(4-2) + 1

N(4) = N(3) + N(2) + 1

N(4) = 7 + 4 + 1                (Using (1) and (2))

∴ N(4) = 12

 

So, minimum number of nodes required to construct AVL tree of height-4 = 12.

 

But given number of nodes = 10 which is less than 12.

Thus, maximum height of AVL tree that can be obtained using 10 nodes = 3.

 

Problem-04:

 

What is the maximum height of any AVL tree with 77 nodes?

 

Solution-

 

For calculating the maximum height of AVL tree with n nodes, we use a recursive relation-

 

N(H) = N(H-1) + N(H-2) + 1

 

Step-01:

 

Substituting H = 2 in the recursive relation, we get-

N(2) = N(2-1) + N(2-2) + 1

N(2) = N(1) + N(0) + 1

N(2) = 2 + 1 + 1                (Using base conditions)

∴ N(2) = 4                          …………(1)

 

So, minimum number of nodes required to construct AVL tree of height-2 = 4.

 

Step-02:

 

Substituting H = 3 in the recursive relation, we get-

N(3) = N(3-1) + N(3-2) + 1

N(3) = N(2) + N(1) + 1

N(3) = 4 + 2 + 1                (Using (1) and base condition)

∴ N(3) = 7                          …………(2)

 

So, minimum number of nodes required to construct AVL tree of height-3 = 7.

 

Step-03:

 

Substituting H = 4 in the recursive relation, we get-

N(4) = N(4-1) + N(4-2) + 1

N(4) = N(3) + N(2) + 1

N(4) = 7 + 4 + 1                (Using (1) and (2))

∴ N(4) = 12                        …………(3)

 

So, minimum number of nodes required to construct AVL tree of height-4 = 12.

 

Step-04:

 

Substituting H = 5 in the recursive relation, we get-

N(5) = N(5-1) + N(5-2) + 1

N(5) = N(4) + N(3) + 1

N(5) = 12 + 7 + 1                (Using (2) and (3))

∴ N(5) = 20                          …………(4)

 

So, minimum number of nodes required to construct AVL tree of height-5 = 20.

 

Step-05:

 

Substituting H = 6 in the recursive relation, we get-

N(6) = N(6-1) + N(6-2) + 1

N(6) = N(5) + N(4) + 1

N(6) = 20 + 12 + 1                (Using (3) and (4))

∴ N(6) = 33                            …………(5)

 

So, minimum number of nodes required to construct AVL tree of height-6 = 33.

 

Step-06:

 

Substituting H = 7 in the recursive relation, we get-

N(7) = N(7-1) + N(7-2) + 1

N(7) = N(6) + N(5) + 1

N(7) = 33 + 20 + 1                (Using (4) and (5))

∴ N(7) = 54                            …………(6)

 

So, minimum number of nodes required to construct AVL tree of height-7 = 54.

 

Step-07:

 

Substituting H = 8 in the recursive relation, we get-

N(8) = N(8-1) + N(8-2) + 1

N(8) = N(7) + N(6) + 1

N(8) = 54 + 33 + 1                (Using (5) and (6))

∴ N(8) = 88                            …………(6)

 

So, minimum number of nodes required to construct AVL tree of height-8 = 88.

 

But given number of nodes = 77 which is less than 88.

Thus, maximum height of AVL tree that can be obtained using 77 nodes = 7.

 

Next Article- Insertion in AVL Tree

 

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AVL Tree Insertion | Insertion in AVL Tree

AVL Tree-

 

Before you go through this article, make sure that you have gone through the previous article on AVL Trees.

 

We have discussed-

  • AVL trees are self-balancing binary search trees.
  • In AVL trees, balancing factor of each node is either 0 or 1 or -1.

 

In this article, we will discuss insertion in AVL tree.

 

Insertion in AVL Tree-

 

Insertion Operation is performed to insert an element in the AVL Tree.

 

To insert an element in the AVL tree, follow the following steps-

  • Insert the element in the AVL tree in the same way the insertion is performed in BST.
  • After insertion, check the balance factor of each node of the resulting tree.

 

Read More- Insertion in Binary Search Tree

 

Now, following two cases are possible-

 

Case-01:

 

  • After the insertion, the balance factor of each node is either 0 or 1 or -1.
  • In this case, the tree is considered to be balanced.
  • Conclude the operation.
  • Insert the next element if any.

 

Case-02:

 

  • After the insertion, the balance factor of at least one node is not 0 or 1 or -1.
  • In this case, the tree is considered to be imbalanced.
  • Perform the suitable rotation to balance the tree.
  • After the tree is balanced, insert the next element if any.

 

Also Read- AVL Tree Properties

 

Rules To Remember-

 

Rule-01:

 

After inserting an element in the existing AVL tree,

  • Balance factor of only those nodes will be affected that lies on the path from the newly inserted node to the root node.

 

Rule-02:

 

To check whether the AVL tree is still balanced or not after the insertion,

  • There is no need to check the balance factor of every node.
  • Check the balance factor of only those nodes that lies on the path from the newly inserted node to the root node.

 

Rule-03:

 

After inserting an element in the AVL tree,

  • If tree becomes imbalanced, then there exists one particular node in the tree by balancing which the entire tree becomes balanced automatically.
  • To re balance the tree, balance that particular node.

 

To find that particular node,

  • Traverse the path from the newly inserted node to the root node.
  • Check the balance factor of each node that is encountered while traversing the path.
  • The first encountered imbalanced node will be the node that needs to be balanced.

 

To balance that node,

  • Count three nodes in the direction of leaf node.
  • Then, use the concept of AVL tree rotations to re balance the tree.

 

PRACTICE PROBLEM BASED ON AVL TREE INSERTION-

 

Problem-

 

Construct AVL Tree for the following sequence of numbers-

50 , 20 , 60 , 10 , 8 , 15 , 32 , 46 , 11 , 48

 

Solution-

 

Step-01: Insert 50

 

 

Step-02: Insert 20

 

  • As 20 < 50, so insert 20 in 50’s left sub tree.

 

 

Step-03: Insert 60

 

  • As 60 > 50, so insert 60 in 50’s right sub tree.

 

 

Step-04: Insert 10

 

  • As 10 < 50, so insert 10 in 50’s left sub tree.
  • As 10 < 20, so insert 10 in 20’s left sub tree.

 

 

Step-05: Insert 8

 

  • As 8 < 50, so insert 8 in 50’s left sub tree.
  • As 8 < 20, so insert 8 in 20’s left sub tree.
  • As 8 < 10, so insert 8 in 10’s left sub tree.

 

 

To balance the tree,

  • Find the first imbalanced node on the path from the newly inserted node (node 8) to the root node.
  • The first imbalanced node is node 20.
  • Now, count three nodes from node 20 in the direction of leaf node.
  • Then, use AVL tree rotation to balance the tree.

 

Following this, we have-

 

 

Step-06: Insert 15

 

  • As 15 < 50, so insert 15 in 50’s left sub tree.
  • As 15 > 10, so insert 15 in 10’s right sub tree.
  • As 15 < 20, so insert 15 in 20’s left sub tree.

 

 

To balance the tree,

  • Find the first imbalanced node on the path from the newly inserted node (node 15) to the root node.
  • The first imbalanced node is node 50.
  • Now, count three nodes from node 50 in the direction of leaf node.
  • Then, use AVL tree rotation to balance the tree.

 

Following this, we have-

 

 

Step-07: Insert 32

 

  • As 32 > 20, so insert 32 in 20’s right sub tree.
  • As 32 < 50, so insert 32 in 50’s left sub tree.

 

 

Step-08: Insert 46

 

  • As 46 > 20, so insert 46 in 20’s right sub tree.
  • As 46 < 50, so insert 46 in 50’s left sub tree.
  • As 46 > 32, so insert 46 in 32’s right sub tree.

 

 

Step-09: Insert 11

 

  • As 11 < 20, so insert 11 in 20’s left sub tree.
  • As 11 > 10, so insert 11 in 10’s right sub tree.
  • As 11 < 15, so insert 11 in 15’s left sub tree.

 

 

Step-10: Insert 48

 

  • As 48 > 20, so insert 48 in 20’s right sub tree.
  • As 48 < 50, so insert 48 in 50’s left sub tree.
  • As 48 > 32, so insert 48 in 32’s right sub tree.
  • As 48 > 46, so insert 48 in 46’s right sub tree.

 

 

To balance the tree,

  • Find the first imbalanced node on the path from the newly inserted node (node 48) to the root node.
  • The first imbalanced node is node 32.
  • Now, count three nodes from node 32 in the direction of leaf node.
  • Then, use AVL tree rotation to balance the tree.

 

Following this, we have-

 

 

This is the final balanced AVL tree after inserting all the given elements.

 

To gain better understanding of AVL Tree Insertion,

Watch this Video Lecture

 

Next Article- Introduction to Hashing

 

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AVL Tree | AVL Tree Example | AVL Tree Rotation

AVL Tree-

 

  • AVL trees are special kind of binary search trees.
  • In AVL trees, height of left subtree and right subtree of every node differs by at most one.
  • AVL trees are also called as self-balancing binary search trees.

 

Also Read- Binary Search Trees

 

Example-

 

Following tree is an example of AVL tree-

 

 

This tree is an AVL tree because-

  • It is a binary search tree.
  • The difference between height of left subtree and right subtree of every node is at most one.

 

Following tree is not an example of AVL Tree-

 

 

This tree is not an AVL tree because-

  • The difference between height of left subtree and right subtree of root node = 4 – 2 = 2.
  • This difference is greater than one.

 

Balance Factor-

 

In AVL tree,

  • Balance factor is defined for every node.
  • Balance factor of a node = Height of its left subtree – Height of its right subtree

 

In AVL tree,

Balance factor of every node is either 0 or 1 or -1.

 

AVL Tree Operations-

 

Like BST Operations, commonly performed operations on AVL tree are-

  1. Search Operation
  2. Insertion Operation
  3. Deletion Operation

 

Also Read- Insertion in AVL Tree

 

After performing any operation on AVL tree, the balance factor of each node is checked.

There are following two cases possible-

 

Case-01:

 

  • After the operation, the balance factor of each node is either 0 or 1 or -1.
  • In this case, the AVL tree is considered to be balanced.
  • The operation is concluded.

 

Case-02:

 

  • After the operation, the balance factor of at least one node is not 0 or 1 or -1.
  • In this case, the AVL tree is considered to be imbalanced.
  • Rotations are then performed to balance the tree.

 

AVL Tree Rotations-

 

Rotation is the process of moving the nodes to make tree balanced.

 

Kinds of Rotations-

 

There are 4 kinds of rotations possible in AVL Trees-

 

 

  1. Left Rotation (LL Rotation)
  2. Right Rotation (RR Rotation)
  3. Left-Right Rotation (LR Rotation)
  4. Right-Left Rotation (RL Rotation)

 

Cases Of Imbalance And Their Balancing Using Rotation Operations-

 

Case-01:

 

 

Case-02:

 

 

Case-03:

 

 

Case-04:

 

 

To gain better understanding about AVL Trees and Rotations,

Watch this Video Lecture

 

Download Handwritten Notes Here-

 

 

Next Article- AVL Tree Properties

 

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