Time Complexity of Basic BST Operations
• Search, Insert, Delete
– These operations visit the nodes along a root-toleaf path
– The number of nodes encountered on unique path
depends on the shape of the tree and the position
of the node in the tree
Worst-case running time of each operation?
O(h) where h is the height of the tree
• The height varies
–A “worst-shape” BST
–A “best-shape” BST
–An “average-shape” BST
Different Shapes of Tree
Balanced BST Can Do better
• BSTs are limited because of their bad worstcase performance O(n). A BST with this worstcase structure is no more efficient than a
regular linked list
• Balanced search trees are trees whose heights
in the worst case is O(lg n)
Balanced BST Can Do better
0
1
2
3
.
10
.
13
.
h-1
If the tree is perfectly balanced, we can store more than 16,000 elements and require at most
14 nodes to be checked to locate an element. How many nodes (in the worst case) would we
have to check in a linked list?
What Does it Mean to Balance a BST?
• Tentative Rule
– Require that the left and right subtrees of the root
node have the same height
We can do better
What Does it Mean to Balance a BST? (cont’d)
• Another Tentative Rule
– Require that every node have left and right
subtrees of the same height
Too restrictive  only perfectly balanced trees of 2k – 1 nodes would satisfy
this criterion
What Does it Mean to Balance BST? (cont’d)
• The Rule
– Require that, for every node, the height of the left
and right subtrees can differ by at most one
Balancing a BST
• There are a number of techniques to properly
balance a binary tree
– Approach 1: take all the elements, place them in
an array, sort them, and then reconstruct the tree
(global) (example, and algorithm)
– Approach 2: constantly restructuring the tree
when new elements arrive or elements are
deleted and lead to an unbalanced tree (i.e., selfbalancing trees)
Approach 1: Reorder Data and Build BST
• When all data arrive, store all data in an array,
sort the array. What is the root of the tree?
– the middle element of the array
• Designate for the root of the BST the middle
element of the array (i.e., the middle element of
the array is the first element inserted into the
BST)
• Continue inserting recursively on the left and
right subarrays until all elements in the array
have been inserted into the BST
• 1 2 3 4 5 6 7 (construct the tree)
Approach 1: Reorder Data and Build BST (cont’d)
• This approach has one serious drawback
– All data must be put in an array before the BST can
be created
– Unsuitable or very inefficient when the BST has to
be used while the data to be included in the BST
are still coming
Approach 2: Self balancing with AVL
Tree
• AVL trees offers another way to balance tree
• AVL trees maintain balance of BSTs while
they are being created via insertions of data
• Tree rebalancing can be performed locally if
only a portion of the tree is affected after
insertion or deletion.
AVL Trees
• AVL (Adel`son-Vel`skii and Landis) tree =
– A BST
– With the property: For every node, the heights
of the left and right subtrees differ at most by
one
– Each node contains a value (-1, 1, 0) indicating
which subtree is "heavier”
Balance Factor
– Each node is marked with a balance factor
indicating which subtree is "heavier”
– Balance Factor: height (right subtree) minus height (left subtree)
– A balanced tree 1
(-1, 1, 0)
-1
0
1
0
1
0
AVL Implementation issues:
– Insert and Delete are modified. They restructure
the tree to make it balanced (if necessary)
Fixing Imbalances
 An imbalance is detected when the height difference between two
subtrees of a node becomes greater than 1 or smaller than -1
 There are two cases of imbalances:
-2
1
-1
0
-2
2
or
2
-1
1
0
CASE 1
Imbalance caused by inserting node
in left subtree of left child or right
subtree of right child (i.e., insertion
occurs on the “outside”)
0
or
0
CASE 2
Imbalance caused by inserting node
in right subtree of left child or left
subtree of right child (i.e., insertion
occurs on the “inside”)
Fixing Imbalances (cont’d)
 Fixing an imbalance is done by rotating the tree
 There are two types of rotation:
 single rotation
 for CASE 1 imbalances
 double rotation
 for CASE 2 imbalances
 consists of two single rotations
 The rotations must always preserve the BST property
Fixing Imbalances: Case 1
node with imbalance
6
D
4
4
C
2
A
B
right rotate
node 4 about 6
2
A
6
B C
D
This is a single right rotation. A single left rotation is symmetric.
Fixing Imbalances: Case 1 (cont’d)
left rotate node Q about P
This is a single left rotation
Fixing Imbalances: Case 2
node with imbalance
node with imbalance
6
6
D
4
2
A
2
4
B
C
STEP 1: left rotate
node 4 about 2
4
D
A
B
C STEP2:
right rotate
A
node 4 about 6
2
6
B C
D
Fixing Imbalances: Case 2 (cont’d)
STEP 1: Right rotate R about Q
STEP 2: Left rotate R about P
Note that P can be part of a larger AVL tree; it
can be a child of some other node in the tree