2. Trees
A very important type of graph in CS is
called a tree:
Real
Tree transformation
L23 2
3. Trees – Their Definition
Let A be a set and let T be a relation on
A. We say that T is a tree if there is a
vertex v0 with the property that there
exists a unique path in T from v0 to
every other vertex in A, but no path
from v0 to v0.
4. Trees – Our Definition
We need a way to describe a tree,
specifically a “rooted” tree.
First, a rooted tree has a single root, v0,
which is a vertex with absolutely no edges
coming into it. (in-degree of v0 = 0)
Every other vertex, v, in the tree has
exactly one path to it from v0. (in-degree
of v = 1)
There may be any number of paths coming
out from any vertex.
Denoted (T,v0) is a rooted tree
5. Definitions
Levels – all of the vertices located n-
edges from v0 are said to be at level n.
Level 0
Level 1
Level 2
Level 3
6. More Definitions
A vertex, v, is considered the parent of
all of the vertices connected to it by
edges leaving v.
A vertex, v, is considered the offspring
of the vertex connected to the single
edge entering v.
A vertex, v, is considered the sibling of
all vertices at the same level with the
same parent.
7. More Definitions
A vertex v2 is considered a descendant
of a vertex v1 if there is a path from v1
to v2.
The height of a tree is the number of
the largest level.
The vertices of a tree that have no
offspring are considered leaves.
If the vertices of a level of a tree can be
ordered from left to right, then the tree
is an ordered tree.
8. More Definitions
If every vertex of a tree has at most n
offspring, then the tree is considered an
n-tree.
If every vertex of a tree with offspring
has exactly n offspring, then the tree is
considered a complete n-tree.
When n=2, this is called a binary tree.
9. Giving Meaning to
Vertices and Edges
Trees implied that a vertex is simply an entity
with parents and offspring much like a family
tree.
What if the position of a vertex relative to its
siblings or the vertex itself represented an
operation.
Examples:
Edges from a vertex represent cases from
a switch statement in software
Vertex represented a mathematical
10. Mathematical Order of
Precedence Represented with
Trees
Consider the equation:
(3 – (2 x)) + ((x – 2) – (3 + x))
Each element is combined with another using
an operator, i.e., this expression can be
broken down into a hierarchy of (a b) where
“” represents an operation used to combine
two elements.
We can use a binary tree to represent this
equation with the elements as the leaves.
12. Positional Tree
A positional tree is an n-tree that
relates the direction/angle an edge
comes out of a vertex to a characteristic
of that vertex. For example:
Yes Maybe Left Right X=0 X=2
No X=1
When n=2, then we have a positional
binary tree.
13. Tree to Convert Base-2 to Base-
10
Starting with the first digit, take the left or right edge
to follow the path to the base-10 value.
First digit 0 1
Second digit 0 1 0 1
3rd
digit 0 1 0 1 0 1 0 1
4th
0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
16. Storing Binary Trees in
Computer
“linked lists”. Each item in the list was
comprised of two components:
Data
Pointer to next item in list
Positional binary trees require two links,
one following the right edge and one
following the left edge. This is referred
to as a “doubly linked list.”
Left Pointer Data Right
17. Represent in computer using
Linked List
+ (2)
– (3) – (8)
3 (4) (5) – (9) + (12)
2 (6) x (7) x (10) 2 (11) 3 (13) x (14)
The numbers in parenthesis represent the index from which
they Can be shown in the linked list
18. Doubly Linked List
Index Left Data Right
1 2 ------- 0
2 3 + 8
3 4 root – 5
4 0 3 0
5 6 7
6 0 2 0
7 0 x 0
8 9 – 12
9 10 – 11
10 0 x 0
11 0 2 0
12 13 + 14
13 0 3 0
14 0 x 0
19. Huffman Code
Depending on the frequency of the
letters occurring in a string, the
Huffman Code assigns patterns of
varying lengths of 1’s and 0’s to
different letters.
These patterns are based on the paths
taken in a binary tree.
A Huffman Code Generator can be
found at:
http://www.inf.puc-
rio.br/~sardinha/Huffman/Huffman.html
21. Terminology
“Visiting” a vertex – the act of
performing a task at a vertex, e.g.,
perform a computation or make a
decision.
“Searching” the tree – the process of
visiting each vertex in a specific order
or path. The term “searching” can be
misleading. Just think of it as
“traversing” the tree.
22. Tree Search
The application of some trees involves
traversing the tree in a methodical pattern so
as to address every vertex.
Our book uses the term “search”, but
sometimes search can imply we’re looking for
a particular vertex. This is not the case.
Example:
Assume we want to compute the average
age, maximum age, and minimum age of all
of the children from five families. (Tree is on
next slide.)
23. Search Example
Neighborhood
families
A. Jones Halls Smith Taylor B. Jones
Katy Tommy Taylor Lori Karen Mike
age 3 age 5 age 1 age 4 age 14 age 6
Phil Lexi Bart Ben
age 8 age 2 age 12 age 2
24. Search Example (continued)
To calculate the average, max, and
min ages for all of the children, we
need to have a method for going
through the tree so that we don’t miss
a child.
By defining a rigorous process, not
only can a human be sure not to miss
a vertex, but also an algorithm can be
defined for a computer
25. A Suggested Process for
Searching a Tree
1.Starting at the root, repeatedly take the
leftmost “untraveled” edge until you arrive at
a leaf which should be a child.
2.Include this child in the average, max, and
min calculations.
3.One at a time, go back up the edges until you
reach a vertex that hasn’t had all of its
outgoing edges traveled.
4.If you get back to the root and cannot find an
untraveled edge, you are done. Otherwise,
return to step 1.
26. Vertices Numbered
in Order of Visits
1 Neighborhood
families
2 5 7 14
11
A. Jones Halls Smith Taylor B. Jones
3 4 8 10 13 16
Katy Tommy Taylor Lori Karen Mike
age 3 age 5 age 1 age 4 age 14 age 6
6 9 12 15
Phil Lexi Bart Ben
age 8 age 2 age 12 age 2
27. Preorder Search
This methodical pattern of traversing a tree is
called a preorder search.
Assume v is the root of a binary positional
tree T.
Each vertex of this tree has at most a left vertex,
vL, and a right vertex, vR.
If either vL or vR have offspring, then they are
subtrees of T, and a search can be performed of
them too.
By viewing a tree this way, then the search
method we described in earlier slides can be
performed using a recursive algorithm applied to
each vertex.
The recursive algorithm is repeatedly applied until
every leaf has been reached.
28. Preorder Search Algorithm
A preorder search of a tree has the
following three steps:
1. Visit the root
2. Search the left subtree if it exists
3. Search the right subtree if it exists
The term “search” in steps 2 and 3
implies that we apply all three steps to
the subtree beginning with step 1.
29. Vertices Visited in Alphabetical
Order Using Preorder Search
A
B H
C E I K
D F G J L
30. Prefix or Polish Form
Binary tree representing: (a – b) × (c + (d ÷ e))
x
– +
a b c ÷
d e
Preorder search produces: × – a b + c ÷ d e
31. Polish Form (continued)
Allows us to write complex arithmetic
expressions without using parenthesis
Expression is evaluated by performing
following steps:
Move left to right until you find a string of
the form Fxy, where F is the symbol for a
binary operation and x and y are numbers.
Evaluate x F y and substitute answer for
string Fxy.
Repeat starting at beginning of string
again.
33. Inorder and Postorder
Searches
Preorder search gets its name from the fact
that the operator that joins two items is
evaluated first, e.g., the binary operation 6 –
4 is visited in the order – 6 4.
Inorder search evaluates the expression as it
is written, e.g., the binary operation 6 – 4 is
visited in the order 6 – 4.
Postorder search evaluates the operator after
the elements are read, e.g., the binary
operation 6 – 4 is visited in the order 6 4 –.
34. Inorder Search Algorithm
An inorder search of a tree has the
following three steps:
1. Search the left subtree if it exists
2. Visit the root
3. Search the right subtree if it exists
35. Postorder Search Algorithm
A postorder search of a tree has the
following three steps:
1. Search the left subtree if it exists
2. Search the right subtree if it exists
3. Visit the root
36. Evaluation of Tree Using
Inorder Search
A
B H
C E I K
D F G J L
Resulting string: DCBFEGAIJHKL
37. Evaluation of Tree Using
Postorder Search
A
B H
C E I K
D F G J L
Resulting string: DCFGEBJILKHA
38. Infix Formrepresenting: (a – b) × (c + (d ÷ e))
Binary tree
x
– +
a b c ÷
d e
Inorder search produces: a – b × c + d ÷ e
Unfortunately, without parenthesis, we can’t do
anything with this expression.
39. Postfix or Reverse Polish Form
Binary tree representing: (a – b) × (c + (d ÷ e))
x
– +
a b c ÷
d e
Inorder search produces: a b – c d e ÷ + ×
40. Reverse Polish Form (continued)
Allows us to write complex arithmetic
expressions without using parenthesis
Expression is evaluated by performing
following steps:
Move left to right until you find a string of
the form xyF, where F is the symbol for a
binary operation and x and y are numbers.
Evaluate x F y and substitute answer for
string xyF.
Repeat starting at beginning of string
again.
42. Converting an Ordered
n-tree to a Positional Binary Tree
An ordered n-tree where some vertices
have more than two offspring can be
converted to a positional binary tree.
This allows easier computer
representation with methods such as
linked lists.
A process exists for this conversion that
works on any finite tree.
43. Process to Convert Ordered n-
tree to Positional Binary Tree
A
Starting at the root, the first or
leftmost offspring of a vertex
remains the leftmost vertex in B C D E
the binary tree
The first sibling to the right
of the leftmost vertex A
becomes the right offspring
of the leftmost vertex B
C
Subsequent siblings
become the right offspring D
in succession until last
sibling is converted.
E
45. Conversion Example
A A
B
B C D E C
F G D
F G H I
E
H
J
K I
J K L
L
Preorder search:
Left tree – ABEFCGJKLHID
Right tree – ABEFCGJKLHID
46. Lets see the City 1 City 2 Mileage
problem. Cleveland Philadelp 400
hi
Cleveland Detroit 200
A small startup airline wants
to provide service to the 5 Cleveland Chicago 350
cities in the table to the right. Cleveland Pittsburg 150
Allowing for multiple
connecting flights, determine Philadelphi Detroit 600
all of the direct flights that Philadelphi Chicago 700
would be needed in order to
service all five cities. Philadelphi Pittsburg 300
Detroit Chicago 300
Source: http://www.usembassy Detroit Pittsburg 300
malaysia.org.my/distance.html
Chicago Pittsburg 450
48. Undirected Tree
An undirected tree is simply the
symmetric closure of a tree.
It is relation that results from a tree
where all edge are made bidirectional,
i.e., there is no defined direction.
L23 48
49. Connected Relation
A relation is connected if for every a
and b in R, there is a path from a to b.
It is easier to see a connected relation
using a digraph than it is to describe in
using words.
A A
C C
B B
E E
D D
Connected Not Connected
50. Spanning Tree of connected
relations
Textbook definition: “If R is a symmetric,
connected relation on a set A, we say that a
tree T on A is a spanning tree for R if T is a
tree with exactly the same vertices as R and
which can be obtained from R by deleting
some edges of R.”
Basically, a undirected spanning tree is one
that connects all n elements of A with n-1
edges.
To make a cycle connecting n elements, more
than n-1 edges will be needed. Therefore,
there are no cycles.
51. Weighted Graph
In the past, we have represented a
undirected graph with unlabeled edges.
It can also be represented with a
symmetric binary matrix. A
C
MT = 0 1 1 0
1 0 1 0 B
1 1 0 1 D
0 0 1 0
52. Weighted Graph (continued)
By giving the edges a numeric value
indicating some parameter in the
relation between two vertices, we can
create a weighted tree.
A 3
C
5
4
7
B
D
53. Weighted Graph (continued)
We can still use matrix notation to represent
a weighted graph. Replace the 1’s used to
represent an edge with the edge’s weight. A
0 indicates no edge.
0 5 3 0
MT = 5 0 4 0
3 4 0 7
0 0 7 0
54. Exercise
(Not drawn to scale)
Detroit
300 600
200 300
Cleveland 400
350
Philadelphia
700 150
300
Chicago 450
Pittsburg
Note R relation has 5 vertices .
55. Minimal Spanning Tree
Assume T represents a spanning tree for
an undirected graph.
The total weight of the spanning tree T is
the sum of all of the weights of all of the
edges of T.
The one(s) with the minimum total weight
are called the minimal spanning
tree(s).
As suggested by the “(s)” in the above
definition, there may be a number of
minimal spanning trees for a particular
undirected graph with the same total
weight.
56. Nearest neighbour of a vertex
A vertex u is nearest neighbour of a
vertex v if u and v are adjacent and no
other vertex is joined to v by an edge of
lesser weight.
L23 56
57. Algorithms for Determining the
Minimal Spanning Tree
There are two algorithms presented in
our textbook for determining the
minimal spanning tree of an undirected
graph that is connected and weighted.
Prim’s Algorithm: process of stepping from
vertex to vertex
Kruskal’s Algoritm: searching through
edges for minimum weights
58. Prim’s Algorithm
Let R be a symmetric, connected
relation with n vertices.
1. Choose a vertex v1 of R. Let V = {v1} and
E = { }.
2. Choose a nearest neighbor vi of V that is
adjacent to vj, vj V, and for which the
edge (vi, vj) does not form a cycle with
members of E. Add vi to V and add (vi, vj)
to E.
3. Repeat Step 2 until |E| = n – 1. Then V
contains all n vertices of R, and E contains
the edges of a minimal spanning tree for R.
59. Prim’s Algorithm in English
The goal is to one at a time include a
new vertex by adding a new edge
without creating a cycle
Pick any vertex to start. From it, pick
the edge with the lowest weight.
As you add vertices, you will add
possible edges to follow to new
vertices.
Pick the edge with the lowest weight
to go to a new vertex without creating
a cycle.
60. Kruskal’s Algorithm
Let R be a symmetric, connected
relation with n vertices and let S =
{e1, e2, e3, …ek} be the set of all
weighted edges of R.
1. Choose an edge e1 in S of least weight.
Let E = {e1}. Replace S with S – {e1}.
2. Select an edge ei in S of least weight that
will not make a cycle with members of E.
Replace E with E {ei} and S with S –
{ei}.
3. Repeat Step 2 until |E| = n – 1.
61. Kruskal’s Algorithm in English
The goal is to one at a time include a
new edge without creating a cycle.
Start by picking the edge with the
lowest weight.
Continue to pick new edges without
creating a cycle. Edges do not
necessarily have to be connected.
Stop when you have n-1 edges as R
have n vertices.
62. Answer: Kruskal to Find MST
(Not drawn to scale)
B D
2
H
3 2
5 E
A C
2
3
4
F G
Sequence of edge selections (D,E), (D,H)(A,C)(A,B)(E,G)(E,F)(C,E)
=2+2+2+3+3+4+5=21,
