Program 5

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Program 5

• Converting an infix expression to postfix
• When evaluating an infix expression, the standard
  order
• of operations are followed
• 1. Expressions in parenthesis
• 2. Multiplication/Division from left to right
• 3. Addition/Subtraction from left to right

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Program 5 converting infix to postfix

• It's easy to see which operation is performed
  first,
• second, etc...with an infix expression
• (4 5) (1 2)
• It becomes a little more difficult with postfix
  expressions
• 4512

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Program 5 converting infix to postfix

• To account for this problem, the different
  operators and parenthesis
• are assigned a priority value
• operator priority
• ( 0
• ) 0
• 1
• - 1
• 2
• / 2
• These priorities will be used in order to help
  determine when to pop
• an item from a stack or when to push an item onto
  a stack.

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Program 5 converting infix to postfix

• Start by putting a set of parenthesis around the
  infix
• expression
• - the left one goes on the stack
• - the right one goes on the end of the infix
  string
• Process the infix expression one character at a
  time,
• from left to right
• - if the character is a digit or letter, put it
  on the end of
• the postfix expression

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Program 5 converting infix to postfix

• Process the infix expression one character at a
  time,
• from left to right
• - if the character is a left parenthesis, push
  it on the
• stack
• - if the character is an operator, pop the stack
  until
• the top value has a smaller priority than the
  current
• character. Each character that is popped off
  should
• be put onto the end of the postfix string.
  Once done
• popping, push the operator onto the stack.

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Program 5 converting infix to postfix

• Process the infix expression one character at a
  time,
• from left to right
• - if the character is a right parenthesis, pop
  the stack
• until the top value is a left parenthesis.
  Each value
• that is popped off should be put onto the end
  of the
• postfix string. Once done popping, pop off
  the left
• parenthesis.

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Program 5 converting infix to postfix

• When implementing in C, two "parallel" stacks
  will
• be used
• 1. The first stack will hold the various
  operators and left parenthesis in the infix
  expression.
• 2. The second stack will hold the priority for
  the various operators and left parenthesis.
• Since these stacks are "parallel", every time one
  of
• them is altered the other must also have the same
  
• alteration applied.

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Program 5 converting infix to postfix

• Push a left parenthesis and its corresponding
  priority on the stacks
• Append a right parenthesis onto the end of the
  infix expression
• While the end of the infix expression has not
  been found
• If the character in the infix expression is a
  letter or digit
• Put it on the end of the postfix expression
• Else if the character in the infix expression
  is a left parenthesis
• Push it and its corresponding priority on the
  stacks
• Else if the character in the infix expression
  is an operator
• Find the priority of the current character in
  the infix expression
• Find the priority of the top item on the
  character stack
• While priority of top item on char stack gt
  priority of current char
• Pop the character and integer stacks
• Put the popped character on the end of the
  postfix expression

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Program 5 converting infix to postfix

• Else if the character in the infix expression is
  a right parenthesis
• Find the top character on the character stack
• While the top character is not a left
  parenthesis
• Pop the character and integer stacks
• Put the popped character on the end of the
  postfix expression
• Find the new top character on the character
  stack
• Endwhile
• Pop the character and integer stacks to
  remove the left parenthesis
• Endif
• Endwhile

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Program 5 Building an expression tree from a
postfix expression

• This process requires a stack, but this one will
  hold tree nodes,
• rather than data.
• The postfix expression will be processed one
  character at a time
• - if the character is a digit, then it is the
  beginning of a constant
• value. Create a tree node that has a data
  field equal to the
• constant value and push it on the stack.
• NOTE Dont forget to ensure that processing
  continues with
• the character AFTER the constant value.

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Program 5 Building an expression tree from a
postfix expression

• The postfix expression will be processed one
  character
• at a time
• - if the character is a letter( ie. a variable),
  create a
• tree node that has a data field equal to the
  letter.
• - if the character is an operator, create a tree
  node
• that has a data field equal to the operator.
  Initialize
• the new nodes right and left children by
  popping the
• stack. Push the new node onto the stack.

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Program 5 Building an expression tree from a
postfix expression

• Once the entire postfix expression is processed,
  there
• will be one node remaining on the stack. It is
  the root of
• the tree.

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C string class

• A standardized string object that can be used in
  place of
• character arrays.
• Multiple Declaration Formats
• string str This invokes the default constructor
  and
• makes str an empty string.
• string str (s) This creates and initializes str
  to contain a copy of
• s, which may be a string or character array
• string str (charAr, n) This creates and
  initializes str to contain a
• copy of the 1st n characters of charAr

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C string class

• Reading a string object
• getline(cin, str)
• This extracts characters from cin and stores
  them in str until a newline character OR end of
  file is reached
• getline(cin, str, delimiter)
• This extracts characters from cin and stores
  them in str until the delimiter is found (not
  stored in str) OR end of file is reached
• cin gtgt str
• This extracts characters from cin and stores
  them in str until whitespace is found OR end of
  file is reached.

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C string class

• Printing a string
• cout ltlt str
• This inserts characters from str into cout
• Assignment Operator Format
• str value
• Assigns a copy of value to str. value may be a
  string, char array, or char

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C string class

• Concatenation Operators
• str value or value str
• returns the result of concatenating str and
  value, which may be a string, char array, or
  character
• str value
• Appends a copy of value to str. value may be a
  string, char array, or character.

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C string class

• Subscript Notation
• strp
• returns a reference to the character stored in
  str at position p
• Relational Operators
• str lt s str lt s
• str gt s str gt s
• str s str ! s
• s may be a string or char array

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C string class

• Useful methods
• str.c_str()
• converts the string object str to a null
  terminated char array
• returns const char
• str.substr(p, n)
• returns a copy of a sub-string n characters
  long, starting at position p

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AVL tree deletions

• As with deletion from a binary search tree, a
  node is deleted from
• an AVL tree using the standard inorder successor
  (predecessor)
• logic for binary search trees.

Deleting 15 using inorder successor logic results
in
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AVL tree deletions

• But, just like insertion, deletion can cause an
  imbalance, which
• will need to be fixed by applying one (or more)
  of the four
• rotations.

Deleting 12 using inorder successor logic results
in
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AVL tree deletions

• What kind of rotation should be performed to fix
  the imbalance?

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AVL tree deletions

• If we identify the parent of the actual node that
  was deleted, then
• - If the left child was deleted, the balance
  factor at the parent
• decreased by 1.
• - If the right child was deleted, the balance
  factor at the parent
• increased by 1.
• The change in balance factor can move up the tree
  from the
• deleted node's parent all the way to the root.
  Therefore, it's
• possible that a node's balance factor could
  become 2 or -2. If
• this happens, then balance needs to be restored.

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AVL tree deletions

• Let A be the node where balance must be restored.
• If the deleted node was in A's right subtree,
  then let B be the root
• of A's left subtree. Then
• - B has balance factor 0 or 1 after deletion --
  then perform
• a single right rotation
• - B has balance factor -1 after deletion -- then
  perform a
• left-right rotation

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AVL tree deletions

• If the deleted node was in A's left subtree, then
  let B
• be the root of A's right subtree. Then
• - B has balance factor 0 or -1 after deletion
• then perform a single left rotation
• - B has balance factor 1 after deletion -- then
  
• perform a right-left rotation

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AVL tree deletions
After a single left rotation
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AVL tree deletions
Delete 9 from the tree
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AVL tree deletions
A is node 5. 9 was deleted from As right
subtree, so B is node 2, which has balance factor
-1.
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AVL tree deletions
Is the tree balanced after the rotation?
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AVL tree deletions

• Unlike insertions, one rotation may not be enough
  to restore balance to the tree.
• To fix the imbalance, find the next node where
  balance factor is bad (call this A)
• - if As balance factor is positive, then let B
  be As left child
• - if Bs left subtree height is larger than Bs
  right subtree height, then
• perform a single right rotation
• - if Bs right subtree height is larger than
  Bs left subtree height, then
• perform a left-right rotation
• - if Bs left and right subtrees have equal
  height, you may perform either
• rotation

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AVL tree deletions

• To fix the imbalance, find the next node where
  balance factor is bad (call
• this A)
• - if As balance factor is negative, then let B
  be As right child
• - if Bs right subtree height is larger than
  Bs left subtree height,
• then perform a single left rotation
• - if Bs left subtree height is larger than Bs
  right subtree height,
• then perform a right-left rotation
• - if Bs left and right subtrees have equal
  height, you may
• perform either rotation

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AVL tree deletions
What type of rotation must be performed to fix
the imbalance?
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AVL tree deletions
A single left rotation with the highlighted
nodesOR
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AVL tree deletions
A right-left rotation with the highlighted nodes
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AVL tree deletions
Result after the single left rotation
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AVL tree deletions
Result after the right-left rotation