Definition of an AVL tree, Insertion of an AVL tree,Height of an AVL tree,Rotations in AVL tree,Representation of AVL tree,Deletion of an AVL tree,AVL search Tree, R0 rotation,R1 rotation,R-1 rotation in AVL tree

3. Definition:
 One of the more popular balanced tree known as
an AVL tree.
 It was introduced in 1962 by Adelson - Velskii
and Landis.
 An empty binary tree is an AVL tree. If T is a
nonempty binary tree with TL and TR as its left
and right subtrees, then T is an AVL tree iff,
(i) TL and TR are AVL trees
(ii) |hL - hR|<=1 where hL and hR are the
heights of TL and TR, respectively.

4. Let Nh be the minimum number of nodes in an
AVL tree of height h.
In the worst case the height of one of the subtrees
is h-1, and the height of the other is h-2.
Both subtrees are also AVL trees.
Nh=Nh-1 + Nh-2 + 1, N0=0, and N1=1
The similarity between this definition for Nh and
the definition of the Fibonacci numbers,
Fn=Fn-1 +Fn-2, F0=0 and F1=1

5.  AVL Trees are normally represented by using
the linked representation scheme for binary
trees. However, to facilitate insertion and
deletion, a balance factor bf is associated with
each node.
 The balance factor bf(x) of a node x is defined
as
height of left subtree of x – height of
right subtree of x
 From the definition of an AVL tree, it follows
that the permissible balance factors are -1, 0
and 1.

6. 30
5 40
35
20
15
12 18
25
30
-1
0
1
0
0
0
0 0
-1
0
The number outside each node is its balance factor
(a) (b)

7. The tree contains nodes with balance factors
other than -1,0 and 1, it is not an AVL tree.
When an insertion into an AVL tree using
algorithm then the results in a search tree that
has one or more nodes with balance factors
other than -1,0 and 1, the resulting tree is
unbalanced.
We can restore balance(i.e., make all balance
factors -1, 0 and 1) by shifting some of the
subtrees of the unbalanced tree

8. 30
5 40
35
32
-2
1
2
0
30
5 35
32 40
-1
0 0
0 0
(a)Immediately after insertion (b) Following rebalancing

9.  Before examining the subtree movement
needed to restore balance, let us make some
observations about the unbalanced tree that
results from an insertion.
 I1 denotes insertion observation 1.
I1: In the unbalanced tree the balance factors
are limited to -2,-1,0,1 and 2.
I2: A node with balance factor 2 had a balance
factor 1 before the insertion.
Similarly, a node with balance factor -2 had a
balance factor -1 before the insertion.

10. I3: The balance factors of only those nodes on the
path from the root to the newly inserted node can
change as a result of the insertion.
I4: Let A denote the nearest ancestor of the newly
inserted node whose balance factor is either -2 or 2.
The balance factor of all nodes on the path from
A to the newly inserted node was 0 prior to the
insertion.

11. 1 0 2
X X X
XL XR
h-1 h-2
XL X’R
h-1 h-1
X’L XR
h h-2
(a) Before insertion (b) After inserting into XR (c) After inserting into XL
Balance factor of X is inside the node
Subtree heights are below subtree names
 For the balance factor to become 0, the insertion must be
made in XR resulting in an X’R of height h-1.
 The height of X’R must increase to h-1 as all balance
factors on the path from X to the newly inserted node were
0 prior to the insertion.
 The height of X remains h, and the balance factors of the
ancestors of X are the same before and after the insertion,
so the tree is balanced.

12. A
0
1
A
A 2
AR
B 1 AR
h
0B
B’L
BRBL B’L BR
h+1
0
A
BR AR
h h h+1 h h h
(a) Before insertion (b) After inserting into BL (c) After LL rotation
h
Balance factors are inside nodes
Subtree heights are below subtree names

13. h
A
0B
2A
B
Cb
-1
0C
B AAR
h
BL BR
hh
BL
h
CL CR
BL CL CR AR
h h
AR
1
(a) Before insertion (b) After inserting into BR (c) After LR rotation
b=0=>bf(B)=bf(A)=0 after rotation
b=1=>bf(B)=0 and bf(A)=-1 after rotation
b=-1=>bf(B)=1 and bf(A)=0 after rotation

14.  The transformations done to remedy LL and RR
imbalances are often called single rotations, while
those done for LR and RL imbalances are called
double rotations.
 The transformation for an LR imbalance can be viewed
as an RR rotation followed by an LL rotation, while
that for an RL imbalance can be viewed as an LL
rotation followed by an RR rotation.
 A single rotation(LL, LR, RR, or RL) is sufficient to
restore balance if the insertion causes imbalance.

15.  Let q be the parent of the node that was physically deleted.
 The node containing this element is deleted and the right-
child pointer from the root diverted to the only child of the
element node.
 The parent of the deleted node is the root, so q is the root.
 The nodes on the path from the root q have changed as a
result of the deletion, this path backward from q toward
the root.
 If the deletion took place from the left subtree of q, bf(q)
decreases by 1, and if it took place from the right subtree
bf(q) increases by 1.

16.  D1 : If the new balance factor of q is 0, its height
has decreased by 1.
 D2 : If the new balance factor of q is either -1 or
1, its height is the same as before the deletion and the
balance factors of its ancestors are unchanged.
 D3 : If the new balance factor of q is either -2 or
2, the tree is unbalanced at q.

17. 2
A’R
BL BR
h h
B
A
h-1
(b) After deletion from AR
-1
1 A
B
BL
h
BR A’R
h h-1
(c) After R0 rotation
1
0
BL BR
h h
AR
h
A
B
(a) Before deletion
An R0 rotation(single rotation)
0

18.  An R0 imbalanced at A is rectified by performing
the rotation. Notice that the height of the shown
subtree was h + 2 before the deletion and is h + 2
after.
 So the balance factors of the remaining nodes on the
path to the root are unchanged.
 How to handle an R1 imbalance. While the pointer
changes are the same as for an R0 imbalance,
 The new balance factors for A and B are different
and the height of the subtree the notation is now h + 1,
which is 1 less than before the deletion.

19. 1
1
2
1
0
0
A
B
BL
h
BR
h-1
AR
h
A
B A’R
h-1
BL BR
h-1h
B
A
BL
h
BR A’R
h-1 h-1
(a) Before deletion (b) After deletion from AR (c) After R1 rotation
An R1 rotation (single rotation)

20.  R1 rotation, we must continue to examine nodes
on the path to the root.
 Unlike the case of an insertion, one rotation
may not suffice to restore balance following a
deletion.
 The number of rotation needed is 0(log n).
 The transformation needed when the imbalance
is of type R- 1.
 The balance factors of A and B following the
rotation depend on the balance factor b of the
right child B.

21. 1
-1
b
2
-1
b
0A
B
BL
h - 1
CL CR
C
AR
h
A
B A’R
h - 1
CBL
h - 1
CL
CR
C
B A
BL
h - 1
CL CR A’R
h - 1
(a) Before deletion (b) After deletion from AR
(c) After R – 1 rotation
b =0=> bf(A) = bf(B) = 0 after rotation
b = 1 => bf(A) = -1 and bf(B) = 0 after rotation
b = -1 => bf(A) = 0 and bf(B) = 1 after rotation
An R – 1 rotation (double rotation)

22.  This rotation leaves behind a
subtree of height h + 1, while the
subtree height prior to the deletion
was h + 2.
 So we need to continue on the
path to the root.
 LL and R1 rotation are identical.
 LL and R0 rotation differ only in
the final balance factors of A and
B and LR and R-1 rotation are
identical.