1. AVL Trees
 AVL Trees
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2. Insertion
 Inserting a node, v, into an AVL tree changes the
heights of some of the nodes in T.
 The only nodes whose heights can increase are
the ancestors of node v.
 If insertion causes T to become unbalanced, then
some ancestor of v would have a height-
imbalance.
 We travel up the tree from v until we find the first
node x such that its grandparent z is unbalanced.
 Let y be the parent of node x.
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3. Insertion (2)
To rebalance the subtree rooted at z, we must
perform a rotation.
44 44
z x
17 78 17 62
y y
50 88 50 78 z
32 32
48 62 x 48 54 88
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4. Rotations
 Rotation is a way of locally reorganizing a BST.
 Let u,v be two nodes such that u=parent(v)
 Keys(T1) < key(v) < keys(T2) < key (u) < keys(T3)
u
v
T3
T1 T2
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5. Insertion
 Insertion happens in subtree T1.
 ht(T1) increases from h to h+1. z
 Since x remains balanced ht(T2)
y T4
is h or h+1 or h+2. h+1 to h+2
 If ht(T2)=h+2 then x is originally x
unbalanced T3
 If ht(T2)=h+1 then ht(x) does not
increase. T1 T2
 Hence ht(T2)=h. h to h+1 h
 So ht(x) increases from h+1 to
h+2.
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6. Insertion(2)
 Since y remains balanced,
h+3 z ht(T3) is h+1 or h+2 or h+3.
 If ht(T3)=h+3 then y is originally
h+2 to h+3 y T4 unbalanced.
h+1  If ht(T3)=h+2 then ht(y) does not
x increase.
h+1 to h+2 T3
 So ht(T3)=h+1.
h+1
T1 T2  So ht(y) inc. from h+2 to h+3.
h to h+1 h  Since z was balanced ht(T4)
is h+1 or h+2 or h+3.
 z is now unbalanced and so
ht(T4)=h+1.
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7. Single rotation
h+3 z
rotation(y,z)
h+2 to h+3 y h+1 h+3 y
T4
x h+2 x z h+2
h+1 to h+2 T3
h+1
T1 T2 T1 T2 T3 T4
h+1 h h+1 h+1
h to h+1 h
The height of the subtree remains the same after
rotation. Hence no further rotations required
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8. Double rotation h+4 z
h+3 z h+3 x T4 h+1
rotation(x,y)
h+2 y T3
h+2 to h+3 y h
T4 h+1
T1 T2
x h+1 to h+2 h+1 h+1
h+1 T1
rotation(x,z)
h to h+1 T2 T3 h
x h+3
Final tree has same height h+2 y z h+2
as original tree. Hence we
need not go further up the T1 T2 T3 T4
h+1 h+1 h h+1
tree.
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9. Restructuring
 The four ways to rotate nodes in an AVL tree, graphically
represented
-Single Rotations:
a=z single rotation b=y
b=y a=z c=x
c=x
T0 T3
T1 T3 T0 T1 T2
T2
c=z single rotation b=y
b=y a=x c=z
a=x
T3 T3
T0 T2 T2 T1 T0 9
T1

10. Restructuring (contd.)
 double rotations:
a=z double rotation b=x
c=y a=z c=y
b=x
T0 T2
T2 T3 T0 T1 T3
T1
c=z double rotation b=x
a=y a=y c=z
b=x
T0 T2
T3 T2 T3 T1 T0
T1
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11. Deletion
 When deleting a node in a BST, we either
delete a leaf or a node with only one child.
 In an AVL tree if a node has only one child
then that child is a leaf.
 Hence in an AVL tree we either delete a
leaf or the parent of a leaf.
 Hence deletion can be assumed to be at a
leaf.
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12. Deletion(2)
 Let w be the node deleted.
 Let z be the first unbalanced node encountered
while travelling up the tree from w. Also, let y be the
child of z with larger height, and let x be the child of
y with larger height.
 We perform rotations to restore balance at the
subtree rooted at z.
 As this restructuring may upset the balance of
another node higher in the tree, we must continue
checking for balance until the root of T is reached
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13. Deletion(3)  Suppose deletion
happens in subtree T4
h+2 z and its ht. reduces from
h+1 y
h to h-1.
T4 h to h-1
 Since z was balanced
h x h or h-1 but is now unbalanced,
T3
ht(y) = h+1.
T1 T2 h-1 or h-2  x has larger ht. than T3
h-1 or h-2 and so ht(x)=h.
 Since y is balanced
ht(T3)= h or h-1
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14. Deletion(4)
h+2 z  Since ht(x)=h, and x is
balanced ht(T1), ht(T2)
h+1 y T4 h to h-1 is h-1 or h-2.
h x  However, both T1 and
T3 h or h-1
T2 cannot have ht. h-2
T1 T2
h-1 or h-2 h-1 or h-2
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15. Single rotation (deletion)
h+2 z
rotation(y,z)
h+1 or h+2
h+1 y T4 h to h-1 y
h or h+1
h x h or h-1
h x z
T3
T1 T2 h-1 or h-2 T1 T2 T3 T4
h-1 or h-2 h-1 or h-2 h-1 or h-2 h or h-1 h-1
After rotation height of subtree might be 1 less than
original height. In that case we continue up the tree
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16. Deletion: another case
h+2 z  As before we can claim
that ht(y)=h+1 and
h+1 y T4 h to h-1
ht(x)=h.
T1
x h  Since y is balanced
h-1 ht(T1) is h or h-1.
T2 T3  If ht(T1) is h then we
h-1 or h-2 h-1 or h-2
would have picked x as
the root of T1.
 So ht(T1)=h-1
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17. Double rotation z h+2
h+1 x T4 h-1
h+2 z rotation(x,y)
h y
T3 h-1 or h-2
h+1 y T4 h to h-1
T1 T2
x h h-1 h-1 or h-2
T1
h-1 rotation(x,z)
T2 T3
h-1 or h-2 h-1 or h-2 x h+1
h y h
Final tree has height less z
than original tree. Hence we
T1 T2 T3 T4 h-1
need to continue up the tree h-1 h-1 or h-2 h-1 or h-2
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18. Running time of insertion & deletion
 Insertion
 We perform rotation only once but might have to
go O(log n) levels to find the unbalanced node.
 So time for insertion is O(log n)
 Deletion
 We need O(log n) time to delete a node.
 Rebalancing also requires O(log n) time.
 More than one rotation may have to be
performed.
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