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Binary Search Trees 1 October 2020 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

  1. Binary Search Trees 1 October 2020 OSU CSE 1

  2. 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) 1 October 2020 OSU CSE 2

  3. 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 1 October 2020 OSU CSE 3

  4. BSTs Are Very General A binary relation on T may be 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 iff that pair is in the set. any type T for which one has defined a total preorder , i.e., a binary relation on T that is total , reflexive , and transitive 1 October 2020 OSU CSE 4

  5. 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 1 October 2020 OSU CSE 5

  6. BSTs Are Very General The binary relation R is reflexive whenever: • 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 1 October 2020 OSU CSE 6

  7. BSTs Are Very General The binary relation R is transitive 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 1 October 2020 OSU CSE 7

  8. 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 1 October 2020 OSU CSE 8

  9. Simplifications • For simplicity in the following illustrations, we use only one kind of example: – T = integer Both these simplifications – The ordering is ≤ 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 1 October 2020 OSU CSE 9

  10. 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 node’s 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 node’s right subtree, then y > x 1 October 2020 OSU CSE 10

  11. The Big Picture x 1 October 2020 OSU CSE 11

  12. The Big Picture Every label y in this tree satisfies y < x x 1 October 2020 OSU CSE 12

  13. The Big Picture Every label y in this tree satisfies y > x x 1 October 2020 OSU CSE 13

  14. And It’s So Everywhere x 1 October 2020 OSU CSE 14

  15. And It’s So Everywhere Every label y in this tree satisfies y < x x 1 October 2020 OSU CSE 15

  16. And It’s So Everywhere Every label y in this tree satisfies y > x x 1 October 2020 OSU CSE 16

  17. Examples of BSTs 7 3 4 5 2 5 1 3 6 9 1 October 2020 OSU CSE 17

  18. Non-Examples of BSTs 1 3 3 5 2 2 5 1 4 4 9 1 October 2020 OSU CSE 18

  19. Non-Examples of BSTs 1 3 3 5 2 2 5 1 4 4 Property 1 is 9 violated here. 1 October 2020 OSU CSE 19

  20. Non-Examples of BSTs 1 3 3 5 2 2 5 1 4 4 Property 1 is 9 violated here. 1 October 2020 OSU CSE 20

  21. Non-Examples of BSTs 1 3 3 5 2 2 5 1 4 4 Property 2 is 9 violated here. 1 October 2020 OSU CSE 21

  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 1 October 2020 OSU CSE 22

  23. 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 . 1 October 2020 OSU CSE 23

  24. Searching for x r 1 October 2020 OSU CSE 24

  25. Searching for x Does x = r ? If so, report that x is in t . If not ... r 1 October 2020 OSU CSE 25

  26. Searching for x Is x < r ? If so, report the result of searching for x in this tree. If not ... r 1 October 2020 OSU CSE 26

  27. Searching for x Then it must be the case that x > r . Report the result of searching for x in this tree. r 1 October 2020 OSU CSE 27

  28. Why It’s Faster Than Linear Search • You need to compare to the root of the tree, and then (only if the root is not what you’re searching for) search either the left or the right subtree—but not both – Compare to linear search, where you might have to look at all the items, which would be equivalent to searching both subtrees 1 October 2020 OSU CSE 28

  29. Example: Searching for 5 8 1 October 2020 OSU CSE 29

  30. Example: Searching for 5 Does 5 = 8 ? No ... 8 1 October 2020 OSU CSE 30

  31. Example: Searching for 5 Is 5 < 8 ? Yes, so report the result of searching for 5 in this tree. 8 1 October 2020 OSU CSE 31

  32. 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, let’s 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 1 October 2020 OSU CSE 32

  33. Example: Searching for 5 8 3 1 October 2020 OSU CSE 33

  34. Example: Searching for 5 Does 5 = 3 ? No ... 8 3 1 October 2020 OSU CSE 34

  35. Example: Searching for 5 Is 5 < 3 ? No ... 8 3 1 October 2020 OSU CSE 35

  36. Example: Searching for 5 Then it must be the case that 5 > 3 . Report the result of searching for 5 in this tree. 8 3 1 October 2020 OSU CSE 36

  37. It’s Another Recursive Call • Let’s continue tracing into calls ... 1 October 2020 OSU CSE 37

  38. Example: Searching for 5 8 3 6 1 October 2020 OSU CSE 38

  39. Example: Searching for 5 Does 5 = 6 ? No ... 8 3 6 1 October 2020 OSU CSE 39

  40. Example: Searching for 5 Is 5 < 6 ? Yes, so report the result of searching for 5 in the (empty) left subtree. 8 3 6 1 October 2020 OSU CSE 40

  41. 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 1 October 2020 OSU CSE 41

  42. Example: Searching for 5 How many nodes did the algorithm visit, and compare labels to 5 ? At worst, how many 8 could it be? 3 6 1 October 2020 OSU CSE 42

  43. Example: Searching for 5 What about in this tree? 8 5 1 October 2020 OSU CSE 43

  44. 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 1 October 2020 OSU CSE 44

  45. Refined Searching for x r 1 October 2020 OSU CSE 45

  46. Refined Searching for x Does x = r ? If so, report that x is in t . If not ... r 1 October 2020 OSU CSE 46

  47. Refined Searching for x Disassemble t into the root and its two subtrees. r 1 October 2020 OSU CSE 47

  48. Refined Searching for x Is x < r ? If so, remember the result of searching for x in this tree. If not ... r 1 October 2020 OSU CSE 48

  49. Refined Searching for x Then it must be the case that x > r . Remember the result of searching for x in this tree. r 1 October 2020 OSU CSE 49

  50. Refined Searching for x Before returning the result of the search, (re)assemble t from its parts. r 1 October 2020 OSU CSE 50

  51. Inserting x • Suppose now you are trying to insert into a BST t the label x (which we assume to be not already in t ; remember that there are no duplicate labels in t ) • There are only two cases to consider: – t is empty – t is non-empty 1 October 2020 OSU CSE 51

  52. Inserting x • Suppose now you are trying to insert into a BST t the label x (which we assume to be not already in t ; remember that there are no duplicate labels in t ) • There are only two cases to consider: – t is empty – t is non-empty Easy: Make x the root of the updated t . 1 October 2020 OSU CSE 52

  53. Inserting x r 1 October 2020 OSU CSE 53

  54. Inserting x There is no reason to ask whether x = r ; why? r 1 October 2020 OSU CSE 54

  55. Inserting x Is x < r ? If so, insert x into this tree. If not ... r 1 October 2020 OSU CSE 55

  56. Inserting x Then it must be the case that x > r . Insert x into this tree. r 1 October 2020 OSU CSE 56

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