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expression trees Oct. 24, 2016 1 Binary tree: each node has at - - PowerPoint PPT Presentation

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COMP 250 Lecture 19 binary trees, expression trees Oct. 24, 2016 1 Binary tree: each node has at most two children. 2 Maximum number of nodes in a binary tree? Height (e.g. 3) 3 Maximum number of nodes in a binary tree? Height

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COMP 250

Lecture 19

binary trees, expression trees

  • Oct. 24, 2016
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Binary tree:

each node has at most two children.

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Maximum number of nodes in a binary tree?

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Height ℎ (e.g. 3)

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Maximum number of nodes in a binary tree?

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Height ℎ (e.g. 3)

𝑜 = 1 + 2 + 4 + 8 + 2ℎ = 2ℎ+1 − 1

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Minimum number of nodes in a binary tree?

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𝑜 = ℎ + 1

Height ℎ (e.g. 3)

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class BTree<T>{ BTNode<T> root; : class BTNode<T>{ T e; BTNode<T> leftchild; BTNode<T> rightchild; : } }

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Binary Tree Traversal (depth first)

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preorder(root){ if (root is not empty){ visit root for each child of root preorder( child ) } }

Rooted tree Binary tree (last lecture)

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Binary Tree Traversal (depth first)

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preorderBT (root){ if (root is not empty){ visit root preorderBT( root.left ) preorderBT( root.right ) } } preorder(root){ if (root is not empty){ visit root for each child of root preorder( child ) } }

Rooted tree Binary tree (last lecture)

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inorderBT (root){ if (root is not empty){ inorderBT(root.left) visit root inorderBT(root.right) } } preorderBT (root){ if (root is not empty){ visit root preorderBT( root.left ) preorderBT( root.right ) } } postorderBT (root){ if (root is not empty){ postorderBT(root.left) postorderBT(root.right) visit root } }

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Example

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a b c d e f g

Pre order: abdecfg In order: debafcg Post order: edbfgca

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Example of binary tree: “Syntax (sub)Tree” of Assignment 2

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statement if boolean then statement else statement end

statement = if boolean then statement else statement end | assignment

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12 statement if boolean then statement else statement end if boolean then statement else statement end if boolean then statement else statement end

The statement nodes form a binary (sub)tree.

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Expression Tree

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+ 3 * 4 2 * + 2 3 4

e.g. 3 + 4 * 2

(3 + 4) * 2 3 + (4 * 2)

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My Windows calculator says 3 + 4 * 2 = 14. Why? (3 + 4) * 2 = 14. Whereas…. if I google “3+4*2”, I get 11. 3 + (4*2) = 11.

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Example of expression tree

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a – b / c + d * e ^ f ^ g

^ is exponentiation We consider binary operators only e.g. we don’t consider 3 + -4 = 3 + (-4) Precedence ordering makes brackets unnecessary. i.e. (a – (b / c)) + (d * (e ^ (f ^ g)))

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Expression Tree

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a – b / c + d * e ^ f ^ g ≡ (a – (b / c)) + (d ∗ (e ^ (f ^ g))) e f g / * a d + b c

Internal nodes are operators. Leaves are numbers.

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Infix, prefix, postfix expressions

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infix: a*b prefix: *ab postfix: ab*

* a b

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Infix, prefix, postfix expressions

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baseExp = variable | integer

  • p = + | - | * | / | ^

inExp = baseExp | inExp op inExp preExp = baseExp | op preExp prefExp postExp = baseExp | postExp postExp

  • p

Use one.

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inorder traversal gives infix expression: a – b / c + d * e ^ f ^ g preorder traversal gives prefix expression: + – a / b c * d ^ e ^ f g postorder traversal gives postfix expression: a b c / - d e f g ^ ^ * +

e f g / * a d + b c If we traverse an expression tree, in which order do we ‘visit’ nodes ?

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If we were given an expression tree, then how would we evaluate the expression ?

e f g / * a d + b c

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If we were given an expression tree, then we could evaluate it using a postorder traversal:

evalExpTree(root){ if (root is a leaf) // root is a number return value else{ // the root is an operator firstOperand = evalExpTree( root.leftchild ) secondOperand = evalExpTree( root.rightchild ) return evaluate(firstOperand, root, secondOperand) } }

However, in practice we are not given an expression tree.

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How to evaluate expressions?

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Infix expressions are awkward to evaluate because of precedence ordering.

ASIDE: One can convert an infix expression to a postfix expression: http://wcipeg.com/wiki/Shunting_yard_algorithm Details omitted here. For your interest only.

We next show how to evaluate a postfix expression using a stack.

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Use a stack to evaluate postfix expression: a b c / - d e f g ^ ^ * +

a ab abc a(bc/) ( a(bc/) - ) ( a(bc/) - ) d ( a(bc/) - ) d e ( a(bc/) - ) d e f ( a(bc/) - ) d e f g ( a(bc/) - ) d e (f g ^) ( a(bc/) - ) d (e (f g ^) ^) ( a(bc/) - ) (d (e (f g ^) ^) * ) (( a(bc/) - ) (d (e (f g ^) ^) * ) +) stack

  • ver

time

e f g / * a d + b c

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Algorithm: Use a stack to evaluate a postfix expression

Let expression be a list of elements. s = empty stack cur = head of expression list while (cur != null){ if ( cur.element is a base expression ) s.push( cur.element ) else{ // cur.element is an operator

  • perand2 = s.pop()
  • perand1 = s.pop()
  • perator = cur.element

// for clarity only s.push( evaluate( operand1, operator, operand2 ) ) } cur = cur.next }

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Prefix expressions called “Polish Notation”

(after Polish logician Jan Lucasewicz 1920’s)

Postfix expressions are called “Reverse Polish notation” (RPN)

Some calculators (esp. Hewlett Packard) require users to input expressions using RPN.

5*4+3 ? 5 <enter> 4 <enter> * <enter> 3 <enter> + <enter> No “=“ symbol needed

  • n keyboard.