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12.txt
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Binary Search Trees
6 May 2013
OSU CSE
1
Faster Searching
The BinaryTree component family can
be used to arrange the labels on binary
tree nodes in a variety of useful ways
A common arrangement of labels, which
supports searching that is much faster
than linear search, is called a binary
search tree (BST)
6 May 2013
OSU CSE
2
BSTs Are Very General
BSTs may be used to search for items of
any type T for which one has defined a
total preorder, i.e., a binary relation on T
that is total, reflexive, and transitive
6 May 2013
OSU CSE
3
A binary relation on T may be
BSTs Are Very General
viewed as a set of ordered pairs of T,
or as a boolean-valued function R of
BSTs may be used to search for items of
two parameters of type T that is true
any type T for which one has defined a
total preorder, i.e., a binary relation on T
that is total, reflexive, and transitive
iff that pair is in the set.
6 May 2013
OSU CSE
4
BSTs Are Very General
The binary relation R is total
whenever:
for all x, y: T
BSTs may be used to search for items of
(R(x, y) or R(y, x))
any type T for which one has defined a
total preorder, i.e., a binary relation on T
that is total, reflexive, and transitive
6 May 2013
OSU CSE
5
BSTs Are Very General
whenever:
The binary relation R is reflexive
BSTs may be used to search for items of
for all x: T (R(x, x))
any type T for which one has defined a
total preorder, i.e., a binary relation on T
that is total, reflexive, and transitive
6 May 2013
OSU CSE
6
The binary relation R is transitive
BSTs Are Very General
whenever:
for all x, y, z: T
BSTs may be used to search for items of
(if R(x, y) and R(y, z)
then R(x, z))
any type T for which one has defined a
total preorder, i.e., a binary relation on T
that is total, reflexive, and transitive
6 May 2013
OSU CSE
7
Simplifications
For simplicity in the following illustrations,
we use only one kind of example:
T = integer
The ordering is
For simplicity (and because of how we will
use BSTs), we assume that no two nodes
in a BST have the same labels
6 May 2013
OSU CSE
8
Simplifications
For simplicity in the following illustrations,
we use only one kind of example:
T = integer
The ordering is
Both these simplifications
are inessential: BSTs are not
For simplicity (and because of how we will
limited to these situations!
use BSTs), we assume that no two nodes
in a BST have the same labels
6 May 2013
OSU CSE
9
BST Arrangement Properties
A binary tree is a BST whenever the
arrangement of node labels satisfies these
two properties:
1. For every node in the tree, if its label is x
and if y is a label in that nodes left subtree,
then y < x
2. For every node in the tree, if its label is x
and if y is a label in that nodes right subtree,
then y > x
6 May 2013
OSU CSE
10
The Big Picture
x
6 May 2013
OSU CSE
11
Every label y in this
tree satisfies
The Big Picture
y < x
x
6 May 2013
OSU CSE
12
The Big Picture
Every label y in this
tree satisfies
y > x
x
6 May 2013
OSU CSE
13
And Its So Everywhere
x
6 May 2013
OSU CSE
14
Every label y in this
And Its So Everywhere
y < x
tree satisfies
x
6 May 2013
OSU CSE
15
And Its So Everywhere
Every label y in this
tree satisfies
y > x
x
6 May 2013
OSU CSE
16
Examples of BSTs
7
4
3
2
5
5
1
3
6
6 May 2013
OSU CSE
9
17
Non-Examples of BSTs
1
3
3
2
2
5
5
1
4
4
6 May 2013
OSU CSE
9
18
Non-Examples of BSTs
1
3
3
2
2
5
5
1
4
4
Property 1 is
violated here.
6 May 2013
OSU CSE
9
19
Non-Examples of BSTs
1
3
3
2
2
5
5
1
4
4
Property 1 is
violated here.
6 May 2013
OSU CSE
9
20
Non-Examples of BSTs
1
3
3
2
2
5
5
1
4
4
Property 2 is
violated here.
6 May 2013
OSU CSE
9
21
Searching for x
Suppose you are trying to find whether
any node in a BST t has the label x
There are only two cases to consider:
t is empty
t is non-empty
6 May 2013
OSU CSE
22
Searching for x
Suppose you are trying to find whether
any node in a BST t has the label x
There are only two cases to consider:
t is empty
t is non-empty
Easy: Report
x is not in t.
6 May 2013
OSU CSE
23
Searching for x
r
6 May 2013
OSU CSE
24
Does x = r?
If so, report that
x is in t.
If not ...
Searching for x
r
6 May 2013
OSU CSE
25
Is x < r?
Searching for x
r
If so, report the
result of searching
for x in this tree.
If not ...
6 May 2013
OSU CSE
26
Then it must be the
case that x > r.
Report the result of
searching for x in
this tree.
Searching for x
r
6 May 2013
OSU CSE
27
Why Its Faster Than Linear Search
You need to compare to the root of the
tree, and then (only if the root is not what
youre searching for) search either the left
or the right subtreebut not both
Compare to linear search, where you might
have to look at all the items, which would be
equivalent to searching both subtrees
6 May 2013
OSU CSE
28
Example: Searching for 5
8
6 May 2013
OSU CSE
29
Example: Searching for 5
Does 5 = 8?
No ...
8
6 May 2013
OSU CSE
30
Example: Searching for 5
Is 5 < 8?
Yes, so report the
result of searching
for 5 in this tree.
8
6 May 2013
OSU CSE
31
Recursion
Searching the left subtree at this point
simply involves making a recursive call to
the method that searches a BST
Against our usual advice about recursion,
lets trace into that call and see what
happens
Why? Because some people, e.g.,
interviewers, may expect you to understand
BSTs without mentioning recursion/induction
6 May 2013
OSU CSE
32
Example: Searching for 5
8
3
6 May 2013
OSU CSE
33
Example: Searching for 5
Does 5 = 3?
No ...
8
3
6 May 2013
OSU CSE
34
Example: Searching for 5
Is 5 < 3?
No ...
8
3
6 May 2013
OSU CSE
35
Then it must be the
case that 5 > 3.
Report the result of
searching for 5 in
this tree.
Example: Searching for 5
8
3
6 May 2013
OSU CSE
36
Its Another Recursive Call
Lets continue tracing into calls ...
6 May 2013
OSU CSE
37
Example: Searching for 5
8
3
6
6 May 2013
OSU CSE
38
Example: Searching for 5
Does 5 = 6?
No ...
8
3
6
6 May 2013
OSU CSE
39
Yes, so report the
result of searching
for 5 in the (empty)
left subtree.
Example: Searching for 5
Is 5 < 6?
8
3
6
6 May 2013
OSU CSE
40
The Recursion Stops Here
Remember, we already noted that when
searching for something in an empty tree,
we can simply report it is not there
No new recursive call results
6 May 2013
OSU CSE
41
Example: Searching for 5
How many nodes
did the algorithm
visit, and compare
At worst, how many
labels to 5?
could it be?
8
3
6
6 May 2013
OSU CSE
42
Example: Searching for 5
What about in
this tree?
8
5
6 May 2013
OSU CSE
43
Wait! How Can This Work?
With the BinaryTree components, there
are no methods to move down the tree
This is why recursion is crucial
To search a subtree, you disassemble the
original tree, search in one of the subtrees,
and then (re)assemble it before returning the
answer
6 May 2013
OSU CSE
44
Refined Searching for x
r
6 May 2013
OSU CSE
45
Refined Searching for x
r
Does x = r?
If so, report that
x is in t.
If not ...
6 May 2013
OSU CSE
46
Disassemble t into
the root and its two
subtrees.
Refined Searching for x
r
6 May 2013
OSU CSE
47
Refined Searching for x
Is x < r?
r
If so, remember the
result of searching
for x in this tree.
If not ...
6 May 2013
OSU CSE
48
Refined Searching for x
Then it must be the
case that x > r.
Remember the
result of searching
for x in this tree.
r
6 May 2013
OSU CSE
49
Before returning the
result of the search,
(re)assemble t
from its parts.