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Data Structures and Algorithms
(CS210/ESO207/ESO211)
Lecture 8
Inventing a new data structure (Binary Search Tree)
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Important Notice
There are basically two ways of introducing a new/innovative solution of a
problem. One way is to just explain it without giving any clue as to how the
person who invented the concept came up with this solution. Another way is
to start from scratch and take a journey of the route which the inventor might
have followed to arrive at the solution. This journey goes through various
hurdles and questions, each hinting towards a better insight into the problem
if we have patience and open mind. Which of these two ways is better ?
I believe that the second way is better and more effective. The current lectureis based on this way. The data structure we shall invent is called a Binary
Search Tree. This is the most fundamental and versatile data structure. We
shall realize this fact many times during the course
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Linkbased Implementation for lists(recap from previous lecture)
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Head
Value
Address of next (or right) node
Singly Linked List
Address of previous (left) node
Doubly Linked List
node
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Doubly Linked List based implementation versus array
based implementation of List
Operation Time Complexity peroperation for arraybased
implementation
Time Complexity peroperation for doubly linked
listbased implementation
IsEmpty(L) O(1) O(1)
Search(x,L) O(n) O(n)
Successor(p,L) O(1) O(1)
Predecessor(p,L) O(1) O(1)
CreateEmptyList(L) O(1) O(1)
Insert(x,p,L) O(n) O(1)
Delete(p,L) O(n) O(1)
MakeListEmpty(L) O(1) O(1)
5Arrays are very rigid
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A fundamental data structure Problem
Maintain a telephone directory
Operations: Search the phone # of a person with name x
Insert a new record (name, phone #,)
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Array based Linked list based
O(n) O(n)
O(1) O(1)
Can we achieve O(log n) bound for
each operation ?
Can we improve it ?
Yes. Keep the array sorted
according to the namesand do
Binary search for x.
Log n
O(n)
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It seems difficult to achieve O(log n) time complexity for insertoperation using Arrays (due to their rigidity which we have
seen). So let us focus on doubly linked lists to explore if it is
possible to achieve an efficient search timeusing them.
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Take a pause for a few minutes to imagine what will
be the structure that will emerge if we pursue our
idea further.
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A new data structure emerges
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A new data structure emergesTo analyze it mathematically, remove irrlevant details
Spend some time over this data structure to see its characteristics.
How does it look like
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Nature is a source of inspiration
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leaves
joints
root
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Nature is a source of inspiration
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Nature is a source of inspiration
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root
leaves
edgesNodes
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Binary Tree: A mathematical model
Definition: A collection of nodes is said to form a binary tree if
1. There is exactly one node with no incoming edge. This node is called the
rootof the tree.
2. Every node other than root node has exactly one incoming edge.
3. Each node has at mosttwo outgoing edges.
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Which of these arenotbinary trees ?
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Binary Search Tree (BST)
Definition:A Binary Tree Tstoring values is said to be Binary Search Tree if for each node vin T
If left(v) NULL, then value(v) > value of every node in subtree(left(v)).
If right(v)NULL, then value(v) < value of every node in subtree(right(v)).
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Look at the similarity of a BSTwith a sorted array.
This will be exploited for searching efficiently an element in a BST.
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Search(T,x)
Searching in a Binary Search Tree
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TSearch(T,33) :
Searching for 33in T.
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Homework 1
Write pseudocode for Insert(T,x) operation similar to
the pseudocode we wrote forSearch(T,x).
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Homework 2
Design an algorithm for the following problem:
Given a sorted array Astoring nelements, build a perfectly
balanced binary search tree storing all elements of Ain O(n)
time.
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A question
Time complexity of Search(T,x)and Insert(T,x) in a Binary
Search Tree T= ??
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O(Height(T))
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Time complexity of Searching and inserting in a
skewed Binary Search tree on nnodes
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T2
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O(n) time !!
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A hurdle on our way
Since the elements may be inserted in arbitrary order into (initially empty)
Binary Search Tree, we may get a skewed tree (with height O(n)). This will
force O(n) time complexity for search and insert operation.
Therefore, we need to modify our algorithm so that the height of tree
remains polylogarithmicof the number of nodes in the tree.
How to do it ? .
Since it is difficult to maintain perfectly balanced BST efficiently (think over
it), we maintain a partially balancedbinary search tree.
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Partially balanced Binary Search Tree
Terminology:Henceforth, sizeof a binary tree would mean the number of
nodes present in it.
Definition:A binary search tree Tis said to be partially balanced at node v, if
subtree(v) consists of either one or two nodes. Or the ratioof the sizeof the subtreesrooted at its two childrenis at most 2.
If none of these conditions hold at v, T is said to be partially imbalanced at
node v.
Definition:A binary search tree T is said to be partially balanced if it is
partially balanced at each node; otherwise T is called partially imbalanced.
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Balancing BST periodically:Preserving O(log n)height after each operation
Each node in T maintains additional field size(v) which is the
number of nodes in the subtree(v).
Keep Search(T,x) operation unchanged.
Modify Insert(T,x) operation as follows:
Carry out normal insert and update the sizefield of appropriate nodes.
If BST T gets partially imbalancedat any node v, perfectly balance
subtree(v).
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Perfectly Balancing subtree at a node v
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v
Size= k
Size> 2kSize differs by at most 1
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Notice that the modified Insert operation described is not
difficult to understand and implement. But it is quite nontrivial
to show that this algorithm solves our problem. In particular,
the following facts are not immediate:
The height of the partially balanced BST on nnodes will be
O(log n).
The total time spent in balancing various partially imbalanced
trees during a sequence of ninsert operations will be small.
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Hopefully you would have now realized that sometimes
analyzing an algorithm is moredifficult than just designing
an algorithm. We shall see many such examples during this
course and the next course (CS345).
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How to analyze this algorithm ?
H(n): maximum height of a partially balanced BST on nnodes.
Question:How to show that H(n) = O(log n) ?
H(1) = 0;H(2) = 1;
H(3) = 1;
Hence H(n) 1 + H(
n)
1 + 1 + H((
)n)
= O( log/n)
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If tree Ton nnodes is partially balanced,
the maximum size of a subtreerooted at
any childof root(T) will be ??2
3( 1)
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Try to achieve the goal mentioned in the
previous slide (hints mentioned will be very
useful). The same goal will be accomplished in
some class next week.
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