The document discusses stacks, which are linear data structures that follow the LIFO (last-in, first-out) principle. Stacks allow elements to be inserted and removed from one end through push and pop operations. Common stack operations are described like push, pop, isEmpty and their functions. Examples of stack operations and conversions between infix, prefix and postfix notations are provided.

1. Stack Data Structure By : Imam M Shofi

2. What is stack? A stack is a limited version of an array. New elements, or nodes as they are often called, can be added to a stack and removed from a stack only from one end. Access system a stack is referred to as a LIFO structure (Last-In First-Out) Some illustrations: stack of satay stack of CDs

3. Stacks operations Push : adds a new node Push(X,S)  add the value X to the TOP of stack Pop : removes a node Pop(S)  removes the TOP node and returns its value IsEmpty : reports whether the stack is empty IsEmpty(S)  report whether the stack S is empty IsFull : reports whether the stack is full IsFull(S)  report whether the stack S is full Initialize : creates/initializes the stack Initialize(S)  create a new empty stack named S Destroy : deletes the contents of the stack (may be implemented by re-initializing the stack) Destroy(S)  deletes the contents of the stack S

4. Illustration/example Operation Stack’s contents TOP value 1. Initialiaze(S) <empty> 0 2. Push(‘a’,S) a 1 3. Push(‘b’,S) a b 2 4. Push(‘c’,S) a b c 3 5. Pop(S) a b 2 6. Push(‘d’,S) a b d 3 7. Push(‘e’,S) a b d e 4 8. Pop(S) a b d 3 9. Pop(S) a b 2 10. Pop(S) a 1

5. Exercise What would the state of the stack be after the following operations: create stack push A onto stack push F onto stack pop item from stack push B onto stack pop item from stack pop item from stack Show the state of the stack and the value of each variable after execution of each of the following statements: A = 5 B = 3 C = 7 (a) create stack push A onto stack push C*C onto stack pop item from stack and store in B push B+A onto stack pop item from stack and store in A pop item from stack and store in B (b) create stack push B onto stack push C onto stack push A onto stack A=B*C push A+C onto stack pop item from stack and store in A pop item from stack and store in B pop item from stack and store in C

6. Converting between notations INFIX to PREFIX : (A+B) – (C*D) Do the first brace: (A+B), the PREFIX is +AB Do the second brace: (C*D), the PREFIX is *CD The end is operator -: +AB - *CD , the PREFIX is -+AB*CD INFIX to POSTFIX : (A+B) – (C*D) Do the first brace: (A+B), the POSTFIX is AB+ Do the second brace: (C*D), the POSTFIX is CD* The end is operator -: AB+ - CD* , the PREFIX is AB+CD*- PREFIX to INFIX : +/*A B C D Find the first operator: *, take 2 operands before the operator ( A and B ), the INFIX is (A*B) Find the second operator: /, take 2 operands before the operator ( A*B and C ), the INFIX is ((A*B)/C) Find the third operator: +, take 2 operands before the operator ( ((A*B)/C) and D ), the INFIX is ((A*B)/C)+D

7. Converting between notations(2) PREFIX to POSTFIX : +/*A B C D Find the first operator: *, take 2 operands before the operator ( A and B ), the POSTFIX is AB* Find the second operator: /, take 2 operands before the operator ( AB* and C ), the POSTFIX is AB*C/ Find the third operator: +, take 2 operands before the operator ( AB*C/ and D ), the POSTFIX is AB*C/D+ POSTFIX to INFIX : ABCD*/- Find the first operator: *, take 2 operands before the operator ( C and D ), the INFIX is (C*D) Find the second operator: /, take 2 operands before the operator ( (C*D) and B ), the INFIX is (B/(C*D) Find the third operator: -, take 2 operands before the operator ( (B/(C*D) and A ), the INFIX is A – (B/(C*D) POSTFIX to PREFIX : ABCD*/- Find the first operator: *, take 2 operands before the operator ( C and D ), the PREFIX is *CD Find the second operator: /, take 2 operands before the operator ( *CD and B ), the PREFIX is /B*CD Find the third operator: -, take 2 operands before the operator ( /B*CD and A ), the PREFIX is -A /B*CD

8. Exercise: Converting Convert these INFIX to PREFIX and POSTFIX : A / B – C / D (A + B) ^ 3 – C * D A ^ (B + C) Convert these PREFIX to INFIX and POSTFIX : + – / A B C ^ D E – + D E / X Y ^ + 2 3 – C D Convert these POSTFIX to INFIX and PREFIX : A B C + – G H + I J / * A B ^ C D + –