Chapter 3, Sections 1 and 2:
Binary Search Trees
AVL Trees
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Ordered Dictionaries
Keys are assumed to come from a total
order.

As opposed to the Ch. 2 stuff where the
keys were not necessarily totally ordered.
New operations:
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

closestKeyBefore(k)
closestElemBefore(k)
closestKeyAfter(k)
closestElemAfter(k)
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Our Theme…
All of the data structures in Chapter 3
are useful for implementing ordered
dictionaries.
Typically, we are looking at variations
on an ordered tree
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
Want searching to be efficient
Need to maintain the properties that make
searching efficient, so inserting and
removing elements can be complicated
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First, a general strategy and a
simple (though not really
efficient) implementation….
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Binary Search
Binary search performs operation findElement(k) on a dictionary
implemented by means of an array-based sequence, sorted by key
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
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similar to looking through a phone book for a name
at each step, the number of candidate items is halved
terminates after O(log n) steps
Example: findElement(7)
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h
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Lookup Table
A lookup table is a dictionary implemented by means of a sorted
sequence
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We store the items of the dictionary in an array-based sequence,
sorted by key
We use an external comparator for the keys
Performance:
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findElement takes O(log n) time, using binary search
insertItem takes O(n) time since in the worst case we have to shift
n items to make room for the new item
removeElement take O(n) time since in the worst case we have to
shift nitems to compact the items after the removal
The lookup table is effective only for dictionaries of small size or
for dictionaries on which searches are the most common
operations, while insertions and removals are rarely performed
(e.g., credit card authorizations)
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Now for the Data Structures
Pretty much all variations on the
binary search tree theme
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Binary Search
Tree
A binary search tree is a
binary tree storing keys
(or key-element pairs)
at its internal nodes and
satisfying the following
property:

Let u, v, and w be three
nodes such that u is in
the left subtree of v and
w is in the right subtree
of v. We have
key(u)  key(v)  key(w)
An inorder traversal of a
binary search trees
visits the keys in
increasing order
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External nodes do not
store items
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Search
Algorithm findElement(k, v)
To search for a key k,
if T.isExternal (v)
we trace a downward
return NO_SUCH_KEY
path starting at the root
if k < key(v)
The next node visited
return findElement(k, T.leftChild(v))
depends on the
else if k = key(v)
outcome of the
return element(v)
comparison of k with
else { k > key(v) }
the key of the current
return findElement(k, T.rightChild(v))
node
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If we reach a leaf, the
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key is not found and we
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return NO_SUCH_KEY
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Example:
1
4 =
findElement(4)
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Insertion (Nice case)
To perform operation
insertItem(k, o), we search
for key k
Assume k is not already in
the tree, and let let w be
the leaf reached by the
search
We insert k at node w and
expand w into an internal
node (i.e. we add two
external children to w)
Example: insert 5
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w
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w
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Insertion (Not so nice case)
w
If k is already in the tree,
and let let w be the node at
which the key k is first
encountered
We call
TreeSearch(k.leftChild(w))
and recursively apply
algorithm to node returned.
Example: insert 2
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w
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1
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2
>
9
4
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2
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Deletion
To perform operation
removeElement(k), we
search for key k
Assume key k is in the tree,
and let let v be the node
storing k
If node v has a leaf child w,
we remove v and w from the
tree with operation
removeAboveExternal(w)
Example: remove 4
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4 v
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w
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Next node in inorder
traversal leftmost node in
subtree rooted at right child
Deletion (cont.)
We consider the case where
the key k to be removed is
stored at a node v whose
children are both internal
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
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we find the internal node w
that follows v in an inorder
traversal
we copy key(w) into node v
we remove node w and its
left child z (which must be a
leaf) by means of operation
removeAboveExternal(z)
Example: remove 3
1
v
3
2
8
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w
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v
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2
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Performance
Consider a dictionary
with n items
implemented by means
of a binary search tree
of height h
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the space used is O(n)
methods findElement ,
insertItem and
removeElement take
O(h) time
The height h is O(n) in
the worst case and
O(log n) in the best
case
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AVL Trees
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AVL Tree Definition
AVL trees are
balanced.
An AVL Tree is a
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binary search tree
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such that for every
internal node v of T,
the heights of the
children of v can
differ by at most 1.
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An example of an AVL tree where the
heights are shown next to the nodes:
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1
n(2)
Height of an AVL Tree
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n(1)
Fact: The height of an AVL tree storing n keys is
O(log n).
Proof: First, we’re trying to bound the height of a
tree with n keys. Let’s come at this the reverse
way, and find the lower bound on the minimum
number of internal nodes, n(h), that a tree with
height h must have.

Claim that n(h) grows exponentially. From this it will be
easy to show that height of tree with n nodes is O(log
n).
So, formally, let n(h) be the minimum number of
internal nodes of an AVL tree of height h.
We easily see that n(1) = 1 and n(2) = 2
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n(2)
Height of an AVL Tree
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n(1)
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For n > 2, an AVL tree of height h with minimal number
of nodes contains the root node, one AVL subtree of
height n-1 and another of height n-2.
That is, n(h) = 1 + n(h-1) + n(h-2)
Knowing n(h-1) > n(h-2), we get n(h) > 2n(h-2). This
indicates that n(h) at least doubles every time we
increase h by two, which intuitively says that n(h) grows
at least exponentially. Let’s show this…
By induction, n(h) > 2n(h-2), n(h) > 4n(h-4), n(h) >
8n(n-6), …, which implies that n(h) > 2in(h-2i)

This holds for any positive integer i such that h-2i is positive.
So, choose i so that we get a base case. That is, we
want i such that either h-2i = 1 or h-2i = 2. I.e. i =
(h/2)-1 or i=(h/2)-(1/2).
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n(2)
Height of an AVL Tree
Thus the i we seek is
n( h) > 2
 h2 1
   1.
h
2
3
n(1)
4
So, we have
h 1
2
n(1 or 2) > 2 .
Taking logarithms: h < 2log n(h) +2
Thus the height of an AVL tree is O(log n)
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Insertion in an AVL Tree
Insertion is as in a binary search tree (may violate
AVL property!)
Always done by expanding an external node.
Example:
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a=y
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Note that only nodes
on path from w to root
can now be unbalanced
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w
before insertion of 54
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Restoring Balance
We use a “search and repair” strategy
Let z be first unbalanced node
encountered on path from w up toward
root
Let y be child of z with higher height
(not y must be an ancestor of w), and x
be child of y with higher height (in case
of tie, go with child that is ancestor of
w)
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Trinode Restructuring
let (a,b,c) be an inorder listing of x, y, z
perform the rotations needed to make b the topmost node of
the three
(other two cases
are symmetrical)
a=z
a=z
case 2: double rotation
(a right rotation about c,
then a left rotation about a)
c=y
b=y
T0
T0
b=x
c=x
T1
T3
b=y
T2
case 1: single rotation
(a left rotation about a)
T1
T3
a=z
T0
b=x
T2
c=x
T1
T2
a=z
T3
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c=y
T1
T2
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T3
Insertion Example, continued
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unbalanced...
T0
T2
T1
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...balanced
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T2
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T3
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Restructuring
(as Single Rotations)
Single Rotations:
a=z
single rotation
b=y
c=x
T0
T1
T3
T2
c =z
a=z
T0
single rotation
b=y
b=y
c=x
T1
T3
T2
b=y
a=x
c =z
a=x
T0
T1
T2
T3
T3
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T1
T0
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Restructuring
(as Double Rotations)
double rotations:
double rotation
a=z
c= y
b=x
a=z
c= y
b=x
T0
T3
T2
T1
T0
T1
double rotation
c=z
a=y
T2
T3
b=x
a=y
c=z
b=x
T3
T2
T0
T3
T2
T1
T0
T1
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Removal in an AVL Tree
Removal begins as in a binary search tree, which
means the node removed will become an empty
external node. Its parent, w, may cause an imbalance.
Example:
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before deletion of 32
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after deletion
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Rebalancing after a Removal
Let z be the first unbalanced node encountered while travelling
up the tree from w. Also, let y be the child of z with the larger
height, and let x be the child of y with the larger height.
We perform restructure(x) to restore balance 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
a=z
w
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c=x
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b=y
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Running Times for
AVL Trees
a single restructure is O(1)
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using a linked-structure binary tree
find is O(log n)
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height of tree is O(log n), no restructures needed
insert is O(log n)
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initial find is O(log n)
Restructuring up the tree, maintaining heights is O(log n)
remove is O(log n)

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initial find is O(log n)
Restructuring up the tree, maintaining heights is O(log n)
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