1. LOGO

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

3. LOGO 3

4. 4 Introduction to Trees  A tree is a connected undirected graph that contains no circuits.  Theorem: There is a unique simple path between any two of its nodes.

5. LOGO 5 Introduction to Trees  A tree is a connected undirected graph that contains no circuits.  Theorem: There is a unique simple path between any 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.

6. LOGO 6 Introduction to Trees  A tree is a connected undirected graph that contains no circuits.  Theorem: There is a unique simple path between any 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.  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 ≥ ? ).

7. LOGO 7 Introduction to Trees  A tree is a connected undirected graph that contains no circuits.  Theorem: There is a unique simple path between any 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.  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 ).

8. LOGO 8 Tree and Forest Examples  A Tree:  A Forest: Leaves in green, internal nodes in brown.

9. LOGO 9 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.

10. LOGO 10 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.

11. LOGO 11 Rooted Tree Examples  Note that a given unrooted tree with n nodes yields n different rooted trees. root Same tree except for choice of root

12. LOGO 12 Rooted-Tree Terminology Exercise  Find the parent, children, siblings, ancestors, & descendants of node f. a b c d e f g h i jk l m n o p q r root

13. LOGO 13 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.

14. LOGO 14 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.

15. LOGO 15 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 ≤ ___.

16. LOGO 16 Ordered Rooted Tree  This is just a rooted tree in which the children of each internal node are ordered.

17. LOGO 17 Ordered Rooted Tree  This is just a rooted tree in which the children of each internal node are ordered.  In ordered binary trees, we can define:  left child, right child  left subtree, right subtree

18. LOGO 18 Ordered Rooted Tree  This is just a rooted tree in which the children of each internal node are ordered.  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.

19. LOGO 19 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?

20. LOGO 20 Some Tree Theorems  Any tree with n nodes has e = n−1 edges.  Proof: Consider removing leaves.

21. LOGO 21 Some Tree Theorems  Any tree with n nodes has e = n−1 edges.  Proof: Consider removing leaves.  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. □

22. LOGO 22 Some Tree Theorems  Any tree with n nodes has e = n−1 edges.  Proof: Consider removing leaves.  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.

23. LOGO 23 Some More Tree Theorems  Definition: 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.

24. LOGO 24 Some More Tree Theorems  Definition: 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 m h leaves in an m-ary tree of height h.  Corollary: An m-ary tree with leaves has height h≥  log m . If m is full and balanced then h=  log m .

25. LOGO 25

26. LOGO 26 Applications of Trees  Binary search trees  A simple data structure for sorted lists

27. LOGO 27 Applications of Trees  Binary search trees  A simple data structure for sorted lists  Decision trees  Minimum comparisons in sorting algorithms

28. LOGO 28 Applications of Trees  Binary search trees  A simple data structure for sorted lists  Decision trees  Minimum comparisons in sorting algorithms  Prefix codes  Huffman coding

29. LOGO 29 Applications of Trees  Binary search trees  A simple data structure for sorted lists  Decision trees  Minimum comparisons in sorting algorithms  Prefix codes  Huffman coding  Game trees

30. LOGO 30 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 a i would instead take Θ(n) worst- case time.

31. LOGO 31 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. 7 312 15915 02811 Example:

32. LOGO 32 Recursive Binary Tree Insert procedure insert(T: binary tree, x: item) v := root[T] if v = null then begin root[T] := x; return “Done” end else if v = x return “Already present” else if x v} return insert(rightSubtree[T], x)

33. LOGO 33 Decision Trees  A 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.

34. LOGO 34 Decision Trees  A 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.

35. LOGO 35 Coin-Weighing Problem  Imagine you have 8 coins, one of 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? ?

36. LOGO 36 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 of weighings thus yields a decision tree with branching factor 3.

37. LOGO 37 Applying the Tree Height Theorem  The 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≥  log m .  Thus the decision tree must have height h ≥  log 3 8  =  1.893…  = 2.  Let’s see if we solve the problem with only 2 weighings…

38. LOGO 38 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.

39. LOGO 39 General Balance Strategy  On 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 − 2  n/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.

40. LOGO 40 Coin Balancing Decision Tree  Here’s what the tree looks like in our case: 123 vs 456 1 vs. 2 left: 123 balanced: 78 right: 456 7 vs. 8 4 vs. 5 L:1 R:2B:3 L:4 R:5B:6 L:7 R:8

41. LOGO 41

42. LOGO Universal 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 k v children, from left to right, with A.1, A.2, A.3, …, A.k v. 42

43. LOGO Universal 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 k v children, from left to right, with A.1, A.2, A.3, …, A.k v. This labeling is called the universal address system of the ordered rooted tree. 43

44. LOGO Universal Address Systems 44

45. LOGO Traversal 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 45

46. LOGO Traversal 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  The three common traversals are:  Preorder traversal  Inorder traversal  Postorder traversal 46

47. LOGO Traversal  Let T be an ordered rooted tree with root r.  Suppose T 1, T 2, …,T n are the subtrees at r from left to right in T. r T1T1 T2T2 TnTn 47

48. LOGO Preorder Traversal r T1T1 T2T2 TnTn Step 1: Visit r Step 2: Visit T 1 in preorder Step 3: Visit T 2 in preorder. Step n+1: Visit T n in preorder 48

49. LOGO Preorder Traversal r T1T1 T2T2 TnTn Step 1: Visit r Step 2: Visit T 1 in preorder Step 3: Visit T 2 in preorder. Step n+1: Visit T n in preorder 49

50. LOGO Preorder Traversal r T1T1 T2T2 TnTn Step 1: Visit r Step 2: Visit T 1 in preorder Step 3: Visit T 2 in preorder. Step n+1: Visit T n in preorder 50

51. LOGO Preorder Traversal r T1T1 T2T2 TnTn Step 1: Visit r Step 2: Visit T 1 in preorder Step 3: Visit T 2 in preorder. Step n+1: Visit T n in preorder 51

52. LOGO Preorder Traversal r T1T1 T2T2 TnTn Step 1: Visit r Step 2: Visit T 1 in preorder Step 3: Visit T 2 in preorder. Step n+1: Visit T n in preorder 52

53. LOGO Example M A R EY P M HJ QT Tree: Visiting sequence: 53

54. LOGO Example AM A R EY P M HJ QT Tree: Visiting sequence: 54

55. LOGO Example AMJ A R EY P M HJ QT Tree: Visiting sequence: 55

56. LOGO Example AYMJ A R EY P M HJ QT Tree: Visiting sequence: 56

57. LOGO Example ARYMJ A R EY P M HJ QT Tree: Visiting sequence: 57

58. LOGO Example ARYMHJ A R EY P M HJ QT Tree: Visiting sequence: 58

59. LOGO Example ARYPMHJ A R EY P M HJ QT Tree: Visiting sequence: 59

60. LOGO Example ARYPMHJQ A R EY P M HJ QT Tree: Visiting sequence: 60

61. LOGO Example ARYPMHJQT A R EY P M HJ QT Tree: Visiting sequence: 61

62. LOGO Example AREYPMHJQT A R EY P M HJ QT Tree: Visiting sequence: 62

63. LOGO The 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. 63

64. LOGO The Preorder Traversal of T 64

65. LOGO The Preorder Traversal of T 65

66. LOGO The Preorder Traversal of T 66

67. LOGO The Preorder Traversal of T 67

68. LOGO Inorder Traversal Step 1: Visit T 1 in inorder Step 2: Visit r Step 3: Visit T 2 in inorder. Step n+1: Visit T n in inorder r T1T1 T2T2 TnTn 68

69. LOGO Inorder Traversal Step 1: Visit T 1 in inorder Step 2: Visit r Step 3: Visit T 2 in inorder. Step n+1: Visit T n in inorder r T1T1 T2T2 TnTn 69

70. LOGO Inorder Traversal Step 1: Visit T 1 in inorder Step 2: Visit r Step 3: Visit T 2 in inorder. Step n+1: Visit T n in inorder r T1T1 T2T2 TnTn 70

71. LOGO Inorder Traversal Step 1: Visit T 1 in inorder Step 2: Visit r Step 3: Visit T 2 in inorder. Step n+1: Visit T n in inorder r T1T1 T2T2 TnTn 71

72. LOGO Inorder Traversal Step 1: Visit T 1 in inorder Step 2: Visit r Step 3: Visit T 2 in inorder. Step n+1: Visit T n in inorder r T1T1 T2T2 TnTn 72

73. LOGO Example J A R EY P M HJ QT Tree: Visiting sequence: 73

74. LOGO Example AJ A R EY P M HJ QT Tree: Visiting sequence: 74

75. LOGO Example AMJ A R EY P M HJ QT Tree: Visiting sequence: 75

76. LOGO Example ARMJ A R EY P M HJ QT Tree: Visiting sequence: 76

77. LOGO Example ARYMJ A R EY P M HJ QT Tree: Visiting sequence: 77

78. LOGO Example ARYPMJ A R EY P M HJ QT Tree: Visiting sequence: 78

79. LOGO Example ARYPMHJ A R EY P M HJ QT Tree: Visiting sequence: 79

80. LOGO Example ARYPMHJQ A R EY P M HJ QT Tree: Visiting sequence: 80

81. LOGO Example ARYPMHJQT A R EY P M HJ QT Tree: Visiting sequence: 81

82. LOGO Example AREYPMHJQT A R EY P M HJ QT Tree: Visiting sequence: 82

83. LOGO The 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. 83

84. LOGO The Inorder Traversal of T 84

85. LOGO The Inorder Traversal of T 85

86. LOGO The Inorder Traversal of T 86

87. LOGO The Inorder Traversal of T 87

88. LOGO Postorder Traversal Step 1: Visit T 1 in postorder Step 2: Visit T 2 in postorder. Step n: Visit T n in postorder Step n+1: Visit r r T1T1 T2T2 TnTn 88

89. LOGO Postorder Traversal Step 1: Visit T 1 in postorder Step 2: Visit T 2 in postorder. Step n: Visit T n in postorder Step n+1: Visit r r T1T1 T2T2 TnTn 89

90. LOGO Postorder Traversal Step 1: Visit T 1 in postorder Step 2: Visit T 2 in postorder. Step n: Visit T n in postorder Step n+1: Visit r r T1T1 T2T2 TnTn 90

91. LOGO Postorder Traversal Step 1: Visit T 1 in postorder Step 2: Visit T 2 in postorder. Step n: Visit T n in postorder Step n+1: Visit r r T1T1 T2T2 TnTn 91

92. LOGO Postorder Traversal Step 1: Visit T 1 in postorder Step 2: Visit T 2 in postorder. Step n: Visit T n in postorder Step n+1: Visit r r T1T1 T2T2 TnTn 92

93. LOGO Example J A R EY P M HJ QT Tree: Visiting sequence: 93

94. LOGO Example AJ A R EY P M HJ QT Tree: Visiting sequence: 94

95. LOGO Example ARJ A R EY P M HJ QT Tree: Visiting sequence: 95

96. LOGO Example ARPJ A R EY P M HJ QT Tree: Visiting sequence: 96

97. LOGO Example ARPJQ A R EY P M HJ QT Tree: Visiting sequence: 97

98. LOGO Example ARPJQT A R EY P M HJ QT Tree: Visiting sequence: 98

99. LOGO Example ARPHJQT A R EY P M HJ QT Tree: Visiting sequence: 99

100. LOGO Example ARYPHJQT A R EY P M HJ QT Tree: Visiting sequence: 100

101. LOGO Example AREYPHJQT A R EY P M HJ QT Tree: Visiting sequence: 101

102. LOGO Example AREYPMHJQT A R EY P M HJ QT Tree: Visiting sequence: 102

103. LOGO The 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. 103

104. LOGO The Postorder Traversal of T 104

105. LOGO The Postorder Traversal of T 105

106. LOGO The Postorder Traversal of T 106

107. LOGO The Postorder Traversal of T 107

108. LOGO Representing Arithmetic Expressions  Complicated arithmetic expressions can be represented by an ordered rooted tree  Internal vertices represent operators  Leaves represent operands 108

109. LOGO Representing 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 subtrees 109

110. LOGO Example (x+y) 2 + (x-3)/(y+2) 110

111. LOGO Example (x+y) 2 + (x-3)/(y+2) + x y 111

112. LOGO Example (x+y) 2 + (x-3)/(y+2) + x y 2  112

113. LOGO Example (x+y) 2 + (x-3)/(y+2) + x y 2  – x 3 113

114. LOGO Example (x+y) 2 + (x-3)/(y+2) + x y 2  – x 3 + y 2 114

115. LOGO Example (x+y) 2 + (x-3)/(y+2) + x y 2  – x 3 + y 2 / 115

116. LOGO Example (x+y) 2 + (x-3)/(y+2) + x y 2  – x 3 + y 2 / + 116

117. LOGO Evaluating Prefix Notation  In 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 117

118. LOGO Example + / + 2 2 2 / – 3 2 + 1 0 118

119. LOGO Example + / + 2 2 2 / – 3 2 + 1 0 119

120. LOGO Example + / + 2 2 2 / – 3 2 + 1 0 + / + 2 2 2 / – 3 2 1 120

121. LOGO Example + / + 2 2 2 / – 3 2 + 1 0 + / + 2 2 2 / – 3 2 1 121

122. LOGO Example + / + 2 2 2 / – 3 2 + 1 0 + / + 2 2 2 / – 3 2 1 + / + 2 2 2 / 1 1 122

123. LOGO Example + / + 2 2 2 / – 3 2 + 1 0 + / + 2 2 2 / – 3 2 1 + / + 2 2 2 / 1 1 123

124. LOGO Example + / + 2 2 2 / – 3 2 + 1 0 + / + 2 2 2 / – 3 2 1 + / + 2 2 2 / 1 1 + / + 2 2 2 1 124

125. LOGO Example + / + 2 2 2 / – 3 2 + 1 0 + / + 2 2 2 / – 3 2 1 + / + 2 2 2 / 1 1 + / + 2 2 2 1 125

126. LOGO Example + / + 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

127. LOGO Example + / + 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

128. LOGO Example + / + 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

129. LOGO Example + / + 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

130. LOGO Example + / + 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 3 130

131. LOGO Postfix Notation (Reverse Polish) Traverse in postorder +  – + / + 2 x y x 3 y 2 131

132. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x +  – + / + 2 x y x 3 y 2 132

133. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x y +  – + / + 2 x y x 3 y 2 133

134. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x + y +  – + / + 2 x y x 3 y 2 134

135. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x + y 2 +  – + / + 2 x y x 3 y 2 135

136. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x + y  2 +  – + / + 2 x y x 3 y 2 136

137. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x + y  2 x +  – + / + 2 x y x 3 y 2 137

138. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x + y  2 x 3 +  – + / + 2 x y x 3 y 2 138

139. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x + y  2 x – 3 +  – + / + 2 x y x 3 y 2 139

140. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x + y  2 x – 3 y +  – + / + 2 x y x 3 y 2 140

141. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x + y  2 x – 3 y 2 +  – + / + 2 x y x 3 y 2 141

142. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x + y  2 x – 3 y + 2 +  – + / + 2 x y x 3 y 2 142

143. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x + y  2 x – 3 / y + 2 +  – + / + 2 x y x 3 y 2 143

144. LOGO Postfix Notation (Reverse Polish) Traverse in postorder x + y  2 + x – 3 / y + 2 +  – + / + 2 x y x 3 y 2 144

145. 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 145

146. LOGO Example 2 2 + 2 / 3 2 – 1 0 + / + 146

147. LOGO Example 2 2 + 2 / 3 2 – 1 0 + / + 147

148. LOGO Example 2 2 + 2 / 3 2 – 1 0 + / + 4 2 / 3 2 – 1 0 + / + 148

149. LOGO Example 2 2 + 2 / 3 2 – 1 0 + / + 4 2 / 3 2 – 1 0 + / + 149

150. LOGO Example 2 2 + 2 / 3 2 – 1 0 + / + 4 2 / 3 2 – 1 0 + / + 2 3 2 – 1 0 + / + 150

151. LOGO Example 2 2 + 2 / 3 2 – 1 0 + / + 4 2 / 3 2 – 1 0 + / + 2 3 2 – 1 0 + / + 151

152. LOGO Example 2 2 + 2 / 3 2 – 1 0 + / + 4 2 / 3 2 – 1 0 + / + 2 3 2 – 1 0 + / + 2 1 1 0 + / + 152

153. LOGO Example 2 2 + 2 / 3 2 – 1 0 + / + 4 2 / 3 2 – 1 0 + / + 2 3 2 – 1 0 + / + 2 1 1 0 + / + 153

154. LOGO Example 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

155. LOGO Example 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

156. LOGO Example 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

157. LOGO Example 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

158. LOGO Example 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

159. LOGO 159 Spanning Trees  You will learn about the following topics in tree later in other courses:  Spanning Trees  Minimum Spanning Trees

160. LOGO Good Luck