Chap 7 binary threaded tree

Chapter 7
Binary Threaded Tree

AVL Tree (Adelson, Velski, Landis)
 AVL tree is a special kind tree in which the balance
factor of each can not be more than 0, 1 or -1.
 It also called as height balanced tree.
 Or we can say that height of two child sub tree of any
node differ at most by one where left and right sub
trees again AVL Tree.
By Prof. Raj Sarode 2
Balance Factor (BF) = Height of the left sub tree -
Height of the Right Sub tree
Or
BF = HL – HR

E.g.
A
B
C
D
BF = 0 - 3 = -3
BF = 0 -2 = - 2
BF = 0 - 1 = -1
BF = 0 - 0 = 0
C
A
B
D
E
BF = 0 - 0 = 0
BF = 0 - 1 = -1
BF = 0 - 0 = 0
BF = 0 - 1 = -1
BF = 0 - 0 = 0
(a) Unbalance BST (b) Balance BST
Note:
BF = -1 i.e. right sub tree is higher than left sub tree or it is called right heavy tree
BF = 1 i.e. left sub tree is higher than right sub tree or it is called as left heavy tree.
BF = 0 i.e. Both sub tree are on same height.

Representation of AVL Tree
AVL Tree can also be represented or implemented as other tree in memory
with an additional field to keep track of balance factor.
The balance factor of node represents the difference of height between the
left sub tree and right sub tree of the node.
Left Info Right BF
Structure of AVL Tree node
Struct Node
{
Node * Left;
Int Info;
Node * Right;
Int BF;
};
Node * root;Building AVL Tree or Height Balance Tree
AVL Tree is constructed by using the same process as applied to BST.
The only thing we need to remember that it follow the property of height
balance tree (BF).
The balance factor of each node can be 0, 1 or -1.
While the insertion of deletion is preformed, the tree become unbalanced and
violate the property of AVL tree that time we have to apply different rotation on
the AVL Tree

AVL Tree Rotation
A
B
C
BF = 0 -2 = - 2
BF = 0 - 1 = -1
BF = 0 - 0 = 0
R R
A
B
CBF = 0 - 0 = 0
BF = 1 - 1 = 0
BF = 0 - 0 = 0
A
B
C
BF = 2 - 0 = 2
BF = 1 - 0 = 1
BF = 0 - 0 = 0
L L
C
B
ABF = 0 - 0 = 0
BF = 1 - 1 = 0
BF = 0 - 0 = 0
1. Left to Left Rotation
2. Right to Right Rotation

AVL Tree Rotation
A
B
C
BF = 0 - 2 = - 2
BF = 1 - 0 = 1
BF = 0 - 0 = 0
R L
A
C
BBF = 0 - 0 = 0
BF = 1 - 1 = 0
BF = 0 - 0 = 0
A
B
C
BF = 2 - 0 = 2
BF = 0 - 1 = -1
BF = 0 - 0 = 0
L R
B
C
ABF = 0 - 0 = 0
BF = 1 - 1 = 0
BF = 0 - 0 = 0
3. Left to Right Rotation
4. Right to Left Rotation

Sr.
No.
I/P Node Tree Rotation
1 Tree is balanced
2 Tree is balanced
3 Tree is balanced
23 BF = 0 - 0 = 023
12
23
12
1
0
82 23
12
1
0
82
0
e.g. 23, 12, 82, 15, 16, 57, 50

Sr.
No.
4 Tree is balanced
5
Now Tree is balanced
15
16
23
12
1
-1
82
0
15
0
23
12
1
-2
82
0
15
-1
16
0
23
12
8215
16
R R
0 0
00
1
e.g. 23, 12, 82, 15, 16, 57, 50

e.g. 23, 12, 82, 15, 16, 57, 50
Sr.
No.
6 Tree is balanced
7
Now Tree is balanced
57
23
12
8215
16 57
0 0 0
0 1
0
50 23
12
8215
16 57
50
0 0
0
0
1
2
-1
23
12 82
15
16
57
50
0 0
0
0 0
0
0
L L

Expression Tree
All arithmetic operation are unary or binary so expression tree are generally binary
Left child represent the left operand.
Right child represent the right operand.
In case of unary operator a node can have only one child.
Root node always contains the operator.
Parenthesis are evaluated first then the exponent, followed by multiplication or
division & then addition or subtraction.
E.g.
A + B
A
+
B

E.g. ( A – B * C ) / ( D + E / F )
B
*
C
Step :1 First parenthesis is evaluated so construct the tree for (A – B * C )
B
*
C
A
-
B * C Then (A - B * C)
Step :2 Second parenthesis is evaluated so construct the tree for (D + E / F )
E
/
F
E
/
F
D
+
E / F Then (D + E / F )

E.g. ( A – B * C ) / ( D + E / F )
Step :3 Result of First parenthesis is divided by second parenthesis
B
*
C
A
-
E
/
F
D
+
/
In-order: (A-B*C)/(D+E/F)
Pre-order: / + D / E F – A * B C
Post- order: D E F / + A B C * - /
Since expression is also binary tree so it can also be traversed in three different ways

B Tree
In previous section we learn BST, AVL Tree and Expression Tree but the problem is
that no. of node is more then height of tree is increased also and will consume
more time for different operation.
So to remove such drawback we introduce M way tree or B Tree having following
properties
1. Each node can contain N-1 key elements, where N is the order of tree
2. Each node can have N branches.
3. The root is either a leaf or it has 2 to n sub tree, where N is the order of B tree.
4. All nodes expect the root and leaves can have minimum N/2 and Maximum N
Branches.
5. All the leaf nodes are on the same level so tree is balanced.
6. When the new node is inserted into full node (with N-1 elements), then the
node is split in to two new node at the same level and key with the median
value is inserted in the parent node in case the parent node is the root, a new
root will be created.
7. Full node condition is called the overflow.

Construct the B-Tree of order 4 for the following elements
1, 6, 8, 2, 9, 12, 15, 7, 18, 3, 4, 20
1
1 6
1 6 8
2 6 81 2
6
81
2
6
81 9
Insert 1
Insert 6
Insert 8
Insert 2
Insert 9

1, 6, 8, 2, 9, 12, 15, 7, 18, 3, 4, 20
2
6
81 9 12
Insert 12
2
6
81 9 12 15
Insert 15
2
6
81
9
12 15
2
6
81
9
12 15
Insert 7
7
2
6
81
9
12 157
Insert 18
18

1, 6, 8, 2, 9, 12, 15, 7, 18, 3, 4, 20
3
6
82
9
12 157
Insert 3
181
4
6
82
9
12 157
Insert 4
181 3
4
6
8
2 9
12 157 181 3

1, 6, 8, 2, 9, 12, 15, 7, 18, 3, 4, 20
Insert 20
4
6
8
2 9
12 157 181 3 20
4
6
8
2 9
12
15
7 181 3 20
4
6
8
2 9
12
15
7 181 3 20 Final B Tree

1, 6, 8, 2, 9, 12, 15, 7, 18, 3, 4, 20
Delete 15
4
6
8
2 9
12
15
7 181 3 20
Final B Tree
4
6
8
2 9
12
18
7 201 3

1, 6, 8, 2, 9, 12, 15, 7, 18, 3, 4, 20
Delete 12
Final B Tree
4
6
8
2 9
12
18
7 201 3
4
6
8
2 9 18
7 201 3

B+ Tree
In B Tree we can access record only in random , we cannot traverse the records
sequentially.
After finding particular record we can not get to get the previous or next record but B+
tree has both properties, random access as well as sequential traversal.
In B+ tree the structure of leaf node & internal node is different
Internal node only store keys and child pointer.
Leaf node store keys & Corresponding data items so data items present only in leaf
node

Threaded Binary Tree
Binary trees have a lot of wasted space: the leaf nodes each have 2 null
pointers
We can use these pointers to help us in in-order traversals
Thread: a pointer to other nodes in the tree for replacing the 0-link.
We have the pointers reference the next node in an in-order traversal;
called threads
We need to know if a pointer is an actual link or a thread, so we keep a
Boolean for each pointer

Thank You

Chap 7 binary threaded tree

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