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Recursion on Trees 7 January 2019 OSU CSE 1 Structure of Trees - PowerPoint PPT Presentation

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Recursion on Trees 7 January 2019 OSU CSE 1 Structure of Trees Two views of a tree: A tree is made up of: A root node A string of zero or more child nodes of the root, each of which is the root of its own tree A tree is

  1. Recursion on Trees 7 January 2019 OSU CSE 1

  2. Structure of Trees • Two views of a tree: – A tree is made up of: • A root node • A string of zero or more child nodes of the root, each of which is the root of its own tree – A tree is made up of: • A root node • A string of zero or more subtrees of the root, each of which is another tree 7 January 2019 OSU CSE 2

  3. Structure of Trees This way of viewing a • Two views of a tree: tree treats it as a – A tree is made up of: collection of nodes. • A root node • A string of zero or more child nodes of the root, each of which is the root of its own tree – A tree is made up of: • A root node • A string of zero or more subtrees of the root, each of which is another tree 7 January 2019 OSU CSE 3

  4. Structure of Trees This way of viewing a • Two views of a tree: tree fully reveals its – A tree is made up of: recursive structure. • A root node • A string of zero or more child nodes of the root, each of which is the root of its own tree – A tree is made up of: • A root node • A string of zero or more subtrees of the root, each of which is another tree 7 January 2019 OSU CSE 4

  5. This way of viewing a R tree treats it as a collection of nodes. K B S A C L H E P T G 7 January 2019 OSU CSE 5

  6. This way of viewing a A tree... tree fully reveals its recursive structure. ... and the subtrees of its root, which are also trees. 7 January 2019 OSU CSE 6

  7. Recursive Algorithms • The “in-your-face” recursive structure of trees (in the second way to view them) allows you to implement some methods that operate on trees using recursion – Indeed, this is sometimes the only sensible way to implement those methods 7 January 2019 OSU CSE 7

  8. XMLTree • The methods for XMLTree are named using the collection-of-nodes view of a tree, because most uses of XMLTree (e.g., the XML/RSS projects) do not need to leverage the recursive structure of trees • But some uses of XMLTree demand that you use the recursive view... 7 January 2019 OSU CSE 8

  9. Example /** * Reports the size of an XMLTree. * ... * @ensures * size = [number of nodes in t] */ private static int size(XMLTree t) {...} 7 January 2019 OSU CSE 9

  10. The number of nodes in this tree, t ... 7 January 2019 OSU CSE 10

  11. ... is 1 (the root) plus the total number of nodes in all the subtrees of the root of t . 7 January 2019 OSU CSE 11

  12. Example private static int size(XMLTree t) { int totalNodes = 1; if (t.isTag()) { for ( int i = 0; i < t.numberOfChildren(); i++) { totalNodes += size(t.child(i)); } } return totalNodes; } 7 January 2019 OSU CSE 12

  13. Example private static int size(XMLTree t) { int totalNodes = 1; if (t.isTag()) { for ( int i = 0; i < t.numberOfChildren(); i++) { totalNodes += size(t.child(i)); } This recursive call } reports the size of a return totalNodes; subtree of the root. } 7 January 2019 OSU CSE 13

  14. Example /** * Reports the height of an XMLTree. * ... * @ensures * height = [height of t] */ private static int height(XMLTree t) {...} 7 January 2019 OSU CSE 14

  15. Example private static int height(XMLTree t) { int maxSubtreeHeight = 0; if (t.isTag()) { for ( int i = 0; i < t.numberOfChildren(); i++) { int subtreeHeight = height(t.child(i)); if (subtreeHeight > maxSubtreeHeight) { maxSubtreeHeight = subtreeHeight; } } } return maxSubtreeHeight + 1; } 7 January 2019 OSU CSE 15

  16. Example This recursive call private static int height(XMLTree t) { reports the height of a int maxSubtreeHeight = 0; subtree of the root. if (t.isTag()) { for ( int i = 0; i < t.numberOfChildren(); i++) { int subtreeHeight = height(t.child(i)); if (subtreeHeight > maxSubtreeHeight) { maxSubtreeHeight = subtreeHeight; } } } return maxSubtreeHeight + 1; } 7 January 2019 OSU CSE 16

  17. Example Why is it a good idea private static int height(XMLTree t) { to store the result of int maxSubtreeHeight = 0; the recursive call in a if (t.isTag()) { variable here? for ( int i = 0; i < t.numberOfChildren(); i++) { int subtreeHeight = height(t.child(i)); if (subtreeHeight > maxSubtreeHeight) { maxSubtreeHeight = subtreeHeight; } } } return maxSubtreeHeight + 1; } 7 January 2019 OSU CSE 17

  18. Expression Trees • There are many other uses for XMLTree • Consider an expression tree , which is a representation of a formula you might type into a Java program or into a calculator, such as: (1 + 3) * 5 – (4 / 2) 7 January 2019 OSU CSE 18

  19. Expression Trees What is the value of this • There are many other uses for XMLTree expression? Computing this value is what we mean by • Consider an expression tree , which is a evaluating the expression. representation of a formula you might type into a Java program or into a calculator, such as: (1 + 3) * 5 – (4 / 2) 7 January 2019 OSU CSE 19

  20. Order of Evaluation • What is the order of evaluation of subexpressions in this expression? (1 + 3) * 5 – (4 / 2) 7 January 2019 OSU CSE 20

  21. Order of Evaluation • What is the order of evaluation of subexpressions in this expression? (1 + 3) * 5 – (4 / 2) • Let’s fully parenthesize it to help: ((1 + 3) * 5) – (4 / 2) 7 January 2019 OSU CSE 21

  22. Order of Evaluation • What is the order of evaluation of subexpressions in this expression? (1 + 3) * 5 – (4 / 2) • Let’s fully parenthesize it to help: ((1 + 3) * 5) – (4 / 2) The fully parenthesized version is based on a convention regarding the precedence of operators (e.g., “ * before – ” in ordinary math). 7 January 2019 OSU CSE 22

  23. Order of Evaluation • What is the order in which the subexpressions in this expression are evaluated? ((1 + 3) * 5) – (4 / 2) First (= 4 ) 7 January 2019 OSU CSE 23

  24. Order of Evaluation • What is the order in which the subexpressions in this expression are evaluated? ((1 + 3) * 5) – (4 / 2) First Second (= 4 ) (= 20 ) 7 January 2019 OSU CSE 24

  25. Order of Evaluation • What is the order in which the subexpressions in this expression are evaluated? ((1 + 3) * 5) – (4 / 2) First Second Third (= 4 ) (= 20 ) (= 2 ) 7 January 2019 OSU CSE 25

  26. Order of Evaluation • What is the order in which the subexpressions in this expression are evaluated? ((1 + 3) * 5) – (4 / 2) First Second Fourth Third (= 4 ) (= 20 ) (= 18 ) (= 2 ) 7 January 2019 OSU CSE 26

  27. Order of Evaluation “Inner-most” parentheses first, but there may be some flexibility in the order of evaluation (e.g., / before + • What is the order in which the would work just as well, but not * subexpressions in this expression are before +, in this expression). evaluated? ((1 + 3) * 5) – (4 / 2) First Second Fourth Third (= 4 ) (= 20 ) (= 18 ) (= 2 ) 7 January 2019 OSU CSE 27

  28. Tree Representation of Expression • Key: Each operand of an operator must be evaluated before that operator can be evaluated 7 January 2019 OSU CSE 28

  29. Tree Representation of Expression • Key: Each operand of an operator must be evaluated before that operator can be evaluated 1 + 2 – 3 * 4 / 5 7 January 2019 OSU CSE 29

  30. Tree Representation of Expression • So, this approach works: – Last operator evaluated is in root node – Each operator’s left and right operands are its two subtrees (i.e., each operator has two subtrees, each of which is a subexpression in the larger expression) 7 January 2019 OSU CSE 30

  31. ((1 + 3) * 5) – (4 / 2) – * / + 5 4 2 1 3 7 January 2019 OSU CSE 31

  32. Evaluation of Expression Trees • To evaluate any expression tree : – If the root is an operator, then first evaluate the expression trees that are its left (first) and right (second) subtrees; then apply that operator to these two values, and the result is the value of the expression represented by the tree – If the root has no subtrees, then it must be an operand, and that operand is the value of the expression represented by the tree 7 January 2019 OSU CSE 32

  33. To evaluate the expression tree rooted here ... – * / + 5 4 2 1 3 7 January 2019 OSU CSE 33

  34. ... first evaluate this expression tree – (= 20 ) ... * / + 5 4 2 1 3 7 January 2019 OSU CSE 34

  35. ... then evaluate this expression tree – (= 2 ) ... * / + 5 4 2 1 3 7 January 2019 OSU CSE 35

  36. ... then apply the operator in the root – (= 18 ). * / + 5 4 2 1 3 7 January 2019 OSU CSE 36

  37. XML Encoding of Expressions • The difference between an operator and an operand can be encoded in XML tags (e.g., "<operator>" and "<operand>") – The specific operator (e.g., "+", "–", "*", "/") can be either an attribute of an operator tag, or its content – Similarly, the value of an operand (e.g., "1", "34723576", etc.) ... • Given such details for a specific XML encoding of expressions, you should be able to evaluate an expression given an XMLTree for its encoding 7 January 2019 OSU CSE 37

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