LOGO. Trees: In these slides: Introduction to trees Applications of trees Tree traversal 2.

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D i s c r e t e S t r u c t u r e s

Instructor: Dr. Ali Movaghar

Sharif University of Technology Fall 1389

In The Name of Allah

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Trees:

In these slides: • Introduction to trees• Applications of trees• Tree traversal

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Introduction to

Trees

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Introduction to TreesA tree is a connected undirected graph that

contains no circuits. Theorem: There is a unique simple path between any

two of its nodes.

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two of its nodes.

A (not-necessarily-connected) undirected graph without simple circuits is called a forest. You can think of it as a set of trees having disjoint sets

of nodes.

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two of its nodes.


of nodes.

A leaf node in a tree or forest is any pendant or isolated vertex. An internal node is any non-leaf vertex (thus it has degree ≥ ? ).

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two of its nodes.


of nodes.

A leaf node in a tree or forest is any pendant or isolated vertex. An internal node is any non-leaf vertex (thus it has degree ≥ 2 ).

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Tree and Forest Examples

A Tree:A Forest:

Leaves in green, internal nodes in brown.

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Rooted Trees

A rooted tree is a tree in which one node has been designated the root. Every edge is (implicitly or explicitly) directed

away from the root.

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Rooted Trees

A rooted tree is a tree in which one node has been designated the root. Every edge is (implicitly or explicitly) directed

away from the root.

Concepts related to rooted trees: Parent, child, siblings, ancestors, descendents, leaf,

internal node, subtree.

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Rooted Tree Examples

Note that a given unrooted tree with n nodes yields n different rooted trees.

root

root

Same tree except

for choiceof root

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Rooted-Tree Terminology Exercise

Find the parent,children, siblings,ancestors, & descendants of node f.

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n-ary trees

A rooted tree is called n-ary if every vertex has no more than n children. It is called full if every internal (non-leaf) vertex

has exactly n children.

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n-ary trees

A rooted tree is called n-ary if every vertex has no more than n children. It is called full if every internal (non-leaf) vertex

has exactly n children.

A 2-ary tree is called a binary tree. These are handy for describing sequences of yes-

no decisions.• Example: Comparisons in binary search algorithm.

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Which Tree is Binary?Theorem: A given rooted tree is a binary tree iff

every node other than the root has degree ≤ ___, and the root has degree ≤ ___.

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Ordered Rooted TreeThis is just a rooted tree in which the children

of each internal node are ordered.

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In ordered binary trees, we can define: left child, right child left subtree, right subtree

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In ordered binary trees, we can define: left child, right child left subtree, right subtree

For n-ary trees with n>2, can use terms like “leftmost”, “rightmost,” etc.

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Trees as Models

Can use trees to model the following: Saturated hydrocarbons Organizational structures Computer file systems

In each case, would you use a rooted or a non-rooted tree?

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Some Tree TheoremsAny tree with n nodes has e = n−1 edges.

Proof: Consider removing leaves.

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A full m-ary tree with i internal nodes has n=mi+1 nodes, and =(m−1)i+1 leaves. Proof: There are mi children of internal nodes, plus the

root. And, = n−i = (m−1)i+1. □

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A full m-ary tree with i internal nodes has n=mi+1 nodes, and =(m−1)i+1 leaves. Proof: There are mi children of internal nodes, plus the

root. And, = n−i = (m−1)i+1. □

Thus, when m is known and the tree is full, we can compute all four of the values e, i, n, and , given any one of them.

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Some More Tree TheoremsDefinition: The level of a node is the length of the

simple path from the root to the node. The height of a tree is maximum node level. A rooted m-ary tree with height h is called balanced if all

leaves are at levels h or h−1.

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Some More Tree TheoremsDefinition: The level of a node is the length of the

simple path from the root to the node. The height of a tree is maximum node level. A rooted m-ary tree with height h is called balanced if all

leaves are at levels h or h−1.

Theorem: There are at most mh leaves in an m-ary tree of height h. Corollary: An m-ary tree with leaves has height

h≥logm . If m is full and balanced then h=logm.

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Applications of

Trees

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Applications of Trees

Binary search trees A simple data structure for sorted lists

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Decision trees Minimum comparisons in sorting algorithms

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Prefix codes Huffman coding

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Prefix codes Huffman coding

Game trees

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Binary Search Trees

A representation for sorted sets of items. Supports the following operations in Θ(log n)

average-case time:• Searching for an existing item.• Inserting a new item, if not already present.

Supports printing out all items in Θ(n) time.

Note that inserting into a plain sequence ai would instead take Θ(n) worst-case time.

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Binary Search Tree Format

Items are stored at individual tree nodes.

We arrange for the tree to always obey this invariant: For every item x,

• Every node in x’s left subtree is less than x.

• Every node in x’s right subtree is greater than x.

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1 5 9 15

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Example:

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Recursive Binary Tree Insert

procedure insert(T: binary tree, x: item)v := root[T]if v = null then begin

root[T] := x; return “Done” endelse if v = x return “Already present”else if x < v then

return insert(leftSubtree[T], x)else {must be x > v}

return insert(rightSubtree[T], x)

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Decision TreesA decision tree represents a decision-making process.

Each possible “decision point” or situation is represented by a node.

Each possible choice that could be made at that decision point is represented by an edge to a child node.

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Decision TreesA decision tree represents a decision-making process.

Each possible “decision point” or situation is represented by a node.

Each possible choice that could be made at that decision point is represented by an edge to a child node.

In the extended decision trees used in decision analysis, we also include nodes that represent random events and their outcomes.

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Coin-Weighing Problem

Imagine you have 8 coins, oneof which is a lighter counterfeit, and a free-beam balance. No scale of weight markings

is required for this problem!

How many weighings are needed to guarantee that the counterfeit coin will be found?

?

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As a Decision-Tree Problem

In each situation, we pick two disjoint and equal-size subsets of coins to put on the scale.

The balance then“decides” whether to tip left, tip right, or stay balanced.

A given sequence ofweighings thus yieldsa decision tree withbranching factor 3.

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Applying the Tree Height TheoremThe decision tree must have at least 8 leaf

nodes, since there are 8 possible outcomes. In terms of which coin is the counterfeit one.

Recall the tree-height theorem, h≥logm. Thus the decision tree must have height

h ≥ log38 = 1.893… = 2.Let’s see if we solve the problem with only 2

weighings…

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General Solution Strategy

The problem is an example of searching for 1 unique particular item, from among a list of n otherwise identical items. Somewhat analogous to the adage of “searching for a needle in haystack.”

Armed with our balance, we can attack the problem using a divide-and-conquer strategy, like what’s done in binary search. We want to narrow down the set of possible locations where the desired item

(coin) could be found down from n to just 1, in a logarithmic fashion. Each weighing has 3 possible outcomes.

Thus, we should use it to partition the search space into 3 pieces that are as close to equal-sized as possible.

This strategy will lead to the minimum possible worst-case number of weighings required.

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General Balance StrategyOn each step, put n/3 of the n coins to be

searched on each side of the scale. If the scale tips to the left, then:

• The lightweight fake is in the right set of n/3 ≈ n/3 coins. If the scale tips to the right, then:

• The lightweight fake is in the left set of n/3 ≈ n/3 coins. If the scale stays balanced, then:

• The fake is in the remaining set of n − 2n/3 ≈ n/3 coins that were not weighed!

Except if n mod 3 = 1 then we can do a little better by weighing n/3 of the coins on each side.

You can prove that this strategy always leads to a balanced 3-ary tree.

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Coin Balancing Decision Tree

Here’s what the tree looks like in our case:

123 vs 456

1 vs. 2

left:123 balanced:

78right:456

7 vs. 84 vs. 5

L:1 R:2 B:3 L:4 R:5 B:6 L:7 R:8

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Tree Traversal

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LOGOUniversal Address Systems

To totally order the vertices of on ordered rooted tree:

1. Label the root with the integer 0.

2. Label its k children (at level 1) from left to right with 1, 2, 3, …, k.

3. For each vertex v at level n with label A, label its kv children, from left to right, with A.1, A.2, A.3, …, A.kv.

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To totally order the vertices of on ordered rooted tree:

1. Label the root with the integer 0.

2. Label its k children (at level 1) from left to right with 1, 2, 3, …, k.

3. For each vertex v at level n with label A, label its kv children, from left to right, with A.1, A.2, A.3, …, A.kv.

This labeling is called the universal address system of the ordered rooted tree.

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LOGOTraversal Algorithms

A traversal algorithm is a procedure for systematically visiting every vertex of an ordered rooted tree An ordered rooted tree is a rooted tree where

the children of each internal vertex are ordered

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LOGOTraversal Algorithms

A traversal algorithm is a procedure for systematically visiting every vertex of an ordered rooted tree An ordered rooted tree is a rooted tree where

the children of each internal vertex are orderedThe three common traversals are:

Preorder traversal Inorder traversal Postorder traversal

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LOGOTraversal

Let T be an ordered rooted tree with root r.Suppose T1, T2, …,Tn are the subtrees at r

from left to right in T.

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LOGOPreorder Traversal

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Step 1: Visit rStep 2: Visit T1 in preorderStep 3: Visit T2 in preorder

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LOGOThe Preorder Traversal of T

In which order does a preorder traversal visit the vertices in the ordered rooted tree T shown to the left?Remember:Visit root, then visit subtrees left to right.

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LOGOInorder Traversal

Step 1: Visit T1 in inorderStep 2: Visit rStep 3: Visit T2 in inorder

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LOGOThe Inorder Traversal of T

In which order does an inorder traversal visit the vertices in the ordered rooted tree T shown to the left?

Remember:Visit leftmost tree, visit root, visit other subtrees left to right.

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LOGOPostorder Traversal

Step 1: Visit T1 in postorderStep 2: Visit T2 in postorder

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LOGOExample

A R YP HJ Q T

A

R

EY

P

M

HJ

Q T

Tree:

Visiting sequence:

100

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LOGOExample

A R EYP HJ Q T

A

R

EY

P

M

HJ

Q T

Tree:

Visiting sequence:

101

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LOGOExample

A R EYP MHJ Q T

A

R

EY

P

M

HJ

Q T

Tree:

Visiting sequence:

102

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LOGOThe Postorder Traversal of T

In which order does a postorder traversal visit the vertices in the ordered rooted tree T shown to the left?

Remember:Visit subtrees left to right, then visit root.

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104

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105

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106

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107

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LOGORepresenting Arithmetic Expressions

Complicated arithmetic expressions can be represented by an ordered rooted tree Internal vertices represent operators Leaves represent operands

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LOGORepresenting Arithmetic Expressions

Complicated arithmetic expressions can be represented by an ordered rooted tree Internal vertices represent operators Leaves represent operands

Build the tree bottom-up Construct smaller subtrees Incorporate the smaller subtrees as part of

larger subtrees109

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LOGOExample

(x+y)2 + (x-3)/(y+2)

110

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LOGOExample

(x+y)2 + (x-3)/(y+2)

+

x y

111

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LOGOExample

(x+y)2 + (x-3)/(y+2)

+

x y

2

112

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LOGOExample

(x+y)2 + (x-3)/(y+2)

+

x y

2

–

x 3

113

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LOGOExample

(x+y)2 + (x-3)/(y+2)

+

x y

2

–

x 3

+

y 2

114

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LOGOExample

(x+y)2 + (x-3)/(y+2)

+

x y

2

–

x 3

+

y 2

/

115

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LOGOExample

(x+y)2 + (x-3)/(y+2)

+

x y

2

–

x 3

+

y 2

/

+

116

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LOGOEvaluating Prefix NotationIn an prefix expression, a binary operator

precedes its two operands

The expression is evaluated right-left

Look for the first operator from the right

Evaluate the operator with the two operands immediately to its right

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0

118

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0

119

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0+ / + 2 2 2 / – 3 2 1

120

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0+ / + 2 2 2 / – 3 2 1

121

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0+ / + 2 2 2 / – 3 2 1+ / + 2 2 2 / 1 1

122

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0+ / + 2 2 2 / – 3 2 1+ / + 2 2 2 / 1 1

123

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0+ / + 2 2 2 / – 3 2 1+ / + 2 2 2 / 1 1+ / + 2 2 2 1

124

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0+ / + 2 2 2 / – 3 2 1+ / + 2 2 2 / 1 1+ / + 2 2 2 1

125

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0+ / + 2 2 2 / – 3 2 1+ / + 2 2 2 / 1 1+ / + 2 2 2 1+ / 4 2 1

126

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0+ / + 2 2 2 / – 3 2 1+ / + 2 2 2 / 1 1+ / + 2 2 2 1+ / 4 2 1

127

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0+ / + 2 2 2 / – 3 2 1+ / + 2 2 2 / 1 1+ / + 2 2 2 1+ / 4 2 1

+ 2 1

128

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0+ / + 2 2 2 / – 3 2 1+ / + 2 2 2 / 1 1+ / + 2 2 2 1+ / 4 2 1

+ 2 1

129

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LOGOExample

+ / + 2 2 2 / – 3 2 + 1 0+ / + 2 2 2 / – 3 2 1+ / + 2 2 2 / 1 1+ / + 2 2 2 1+ / 4 2 1

+ 2 1

3130

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LOGOPostfix Notation (Reverse Polish)

• Traverse in postorder

+

– +

/

+ 2

x y x 3 y 2

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x

+

– +

/

+ 2

x y x 3 y 2

132

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x y

+

– +

/

+ 2

x y x 3 y 2

133

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x +y

+

– +

/

+ 2

x y x 3 y 2

134

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x +y 2

+

– +

/

+ 2

x y x 3 y 2

135

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x +y 2

+

– +

/

+ 2

x y x 3 y 2

136

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x +y 2 x

+

– +

/

+ 2

x y x 3 y 2

137

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x +y 2 x 3

+

– +

/

+ 2

x y x 3 y 2

138

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x +y 2 x –3

+

– +

/

+ 2

x y x 3 y 2

139

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x +y 2 x –3 y

+

– +

/

+ 2

x y x 3 y 2

140

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x +y 2 x –3 y 2

+

– +

/

+ 2

x y x 3 y 2

141

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x +y 2 x –3 y +2

+

– +

/

+ 2

x y x 3 y 2

142

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x +y 2 x –3 /y +2

+

– +

/

+ 2

x y x 3 y 2

143

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x +y 2 +x –3 /y +2

+

– +

/

+ 2

x y x 3 y 2

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LOGO

In an postfix expression, a binary operator follows its two operands

The expression is evaluated left-right

Look for the first operator from the left

Evaluate the operator with the two operands immediately to its left

Evaluating Postfix Notation

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LOGOExample

2 2 + 2 / 3 2 – 1 0 + / +

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LOGOExample

2 2 + 2 / 3 2 – 1 0 + / +

147

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LOGOExample

2 2 + 2 / 3 2 – 1 0 + / +

4 2 / 3 2 – 1 0 + / +

148

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LOGOExample

2 2 + 2 / 3 2 – 1 0 + / +

4 2 / 3 2 – 1 0 + / +

149

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LOGOExample

2 2 + 2 / 3 2 – 1 0 + / +

4 2 / 3 2 – 1 0 + / +

2 3 2 – 1 0 + / +

150

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LOGOExample

2 2 + 2 / 3 2 – 1 0 + / +

4 2 / 3 2 – 1 0 + / +

2 3 2 – 1 0 + / +

151

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LOGOExample

2 2 + 2 / 3 2 – 1 0 + / +

4 2 / 3 2 – 1 0 + / +

2 3 2 – 1 0 + / +

2 1 1 0 + / +

152

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LOGOExample

2 2 + 2 / 3 2 – 1 0 + / +

4 2 / 3 2 – 1 0 + / +

2 3 2 – 1 0 + / +

2 1 1 0 + / +

153

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LOGOExample

2 2 + 2 / 3 2 – 1 0 + / +

4 2 / 3 2 – 1 0 + / +

2 3 2 – 1 0 + / +

2 1 1 0 + / +

2 1 1 / +

154

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LOGOExample

2 2 + 2 / 3 2 – 1 0 + / +

4 2 / 3 2 – 1 0 + / +

2 3 2 – 1 0 + / +

2 1 1 0 + / +

2 1 1 / +

155

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LOGOExample

2 2 + 2 / 3 2 – 1 0 + / +

4 2 / 3 2 – 1 0 + / +

2 3 2 – 1 0 + / +

2 1 1 0 + / +

2 1 1 / +

2 1 +

156

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LOGOExample

2 2 + 2 / 3 2 – 1 0 + / +

4 2 / 3 2 – 1 0 + / +

2 3 2 – 1 0 + / +

2 1 1 0 + / +

2 1 1 / +

2 1 +

157

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LOGOExample

3

2 2 + 2 / 3 2 – 1 0 + / +

4 2 / 3 2 – 1 0 + / +

2 3 2 – 1 0 + / +

2 1 1 0 + / +

2 1 1 / +

2 1 +

158

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LOGO

159

Spanning Trees

You will learn about the following topics in tree later in other courses: Spanning Trees Minimum Spanning Trees

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LOGO

LOGO

Any Question?Good Luck

LOGO. Trees: In these slides: Introduction to trees Applications of trees Tree traversal 2.

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Transcript of LOGO. Trees: In these slides: Introduction to trees Applications of trees Tree traversal 2.