Infix to Postfix Conversion Using Stack in DSA Java - Execution Trace

Concept Flow - Infix to Postfix Conversion Using Stack

Read next symbol from infix

↓

Is symbol operand?

Yes→Add operand to postfix

No↓

Is symbol '('?

Yes→Push '(' on stack

No↓

Is symbol ')' ?

Yes→Pop and add to postfix until '(' found

No↓

While stack top has higher or equal precedence

↓

Pop from stack to postfix

↓

Push current operator on stack

↓

Repeat until infix ends

↓

Pop all remaining stack to postfix

↓

Done

The flow reads each symbol from the infix expression, decides if it's an operand or operator, manages the stack accordingly, and builds the postfix expression step-by-step.

Execution Sample

DSA Java

infix = "A+B*C"
stack = []
postfix = ""
for symbol in infix:
  if symbol.isalnum():
    postfix += symbol
  else:
    # manage stack according to precedence
    pass
# pop all stack to postfix

Converts infix expression 'A+B*C' to postfix using a stack by processing each symbol.

Execution Table

Step	Symbol	Operation	Stack State	Postfix Expression	Explanation
1	A	Operand: add to postfix	[]	A	A is operand, add to postfix
2	+	Operator: stack empty, push '+'	[+]	A	Push '+' on stack
3	B	Operand: add to postfix	[+]	AB	B is operand, add to postfix
4	*	Operator: '*' has higher precedence than '+'; push '*'	[+, *]	AB	Push '*' on stack
5	C	Operand: add to postfix	[+, *]	ABC	C is operand, add to postfix
6	End	Pop all from stack to postfix	[]	ABC*+	Pop '*' then '+' to postfix
7	End	Done	[]	ABC*+	Final postfix expression

💡 All symbols processed and stack emptied, postfix expression complete

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	Final
stack	[]	[]	[+]	[+]	[+, *]	[+, *]	[]	[]
postfix	""	"A"	"A"	"AB"	"AB"	"ABC"	"ABC*+"	"ABC*+"

Key Moments - 3 Insights

Why do we push '*' on the stack even when '+' is already there?

Because '*' has higher precedence than '+', so we push '*' without popping '+' (see Step 4 in execution_table).

Why do we pop all remaining operators from the stack at the end?

To add all operators left in stack to postfix expression after processing all symbols (see Step 6).

Why do operands go directly to postfix without stack?

Operands appear in the output in the same order as input, so we add them directly (see Steps 1, 3, 5).

Visual Quiz - 3 Questions

Look at the execution_table, what is the stack state after processing symbol '*'?

A [+, *]

B [+]

C []

D [*]

💡 Hint

Check Step 4 in execution_table for stack after '*' is processed

At which step does the postfix expression become 'ABC'?

A Step 3

B Step 4

C Step 5

D Step 6

💡 Hint

Look at postfix expression column in execution_table

If we change '+' to '-' in the infix, how does the final postfix change?

A It becomes 'ABC*-'

B It becomes 'ABC*+'

C It becomes 'AB+C*'

D '-*CBA' semoceb tI

💡 Hint

Changing '+' to '-' changes the operator in postfix; check final postfix in execution_table

Concept Snapshot

Infix to Postfix Conversion Using Stack:
- Read infix left to right
- Operands go directly to postfix
- '(' pushed on stack
- ')' pops until '('
- Operators pushed after popping higher/equal precedence
- Pop all stack at end
- Result: postfix expression

Full Transcript

This visual trace shows how to convert an infix expression like 'A+B*C' to postfix using a stack. We read each symbol: operands go directly to the postfix output. Operators are pushed on the stack, but before pushing, we pop operators from the stack that have higher or equal precedence. Parentheses are handled specially: '(' is pushed, and ')' causes popping until '(' is found. After processing all symbols, we pop any remaining operators from the stack to the postfix expression. The execution table shows each step, stack state, and postfix built so far. The variable tracker follows stack and postfix changes. Key moments clarify why operators are pushed or popped. The quiz tests understanding of stack states and postfix formation. This method ensures correct operator order in postfix form for easy evaluation later.

Practice

1. What is the main purpose of using a stack in converting an infix expression to postfix?

easy

A. To temporarily hold operators and manage their precedence

B. To store operands permanently

C. To reverse the order of operands

D. To count the number of operators

Solution

1. Step 1: Understand the role of stack in infix to postfix

   The stack is used to hold operators temporarily while processing the expression to respect operator precedence and associativity.

2. Step 2: Differentiate stack use from other data structures

   Operands are output immediately, not stored in stack. Stack is not used for counting or reversing operands.

3. Final Answer:

   To temporarily hold operators and manage their precedence -> Option A

4. Quick Check:

   Stack holds operators = A [OK]

Hint: Stack holds operators to manage precedence during conversion [OK]

Common Mistakes:

• Thinking stack stores operands
• Confusing stack with queue
• Believing stack reverses operands

2. Which of the following is the correct way to push an operator onto the stack in Java?

easy

A. stack.push(operator);

B. stack.insert(operator);

C. stack.add(operator);

D. stack.put(operator);

Solution

1. Step 1: Recall Java Stack method for adding elements

   Java's Stack class uses the push() method to add elements to the top of the stack.

2. Step 2: Identify incorrect methods

   Methods add(), insert(), and put() are not standard for Stack push operation in Java.

3. Final Answer:

   stack.push(operator); -> Option A

4. Quick Check:

   Use push() to add to stack [OK]

Hint: Use push() to add elements to Java stack [OK]

Common Mistakes:

• Using add() instead of push()
• Using non-existent methods like insert()
• Confusing stack with other collections

3. What is the postfix expression for the infix expression A + B * C?

medium

A. AB+C*

B. ABC*+

C. A+BC*

D. *ABC+

Solution

1. Step 1: Apply operator precedence in infix to postfix

   Multiplication (*) has higher precedence than addition (+), so B*C is evaluated first.

2. Step 2: Convert expression step-by-step

   Postfix: operands B and C followed by *, then operand A followed by + gives "ABC*+".

3. Final Answer:

   ABC*+ -> Option B

4. Quick Check:

   Multiplication before addition = ABC*+ [OK]

Hint: Multiply before add: operands then operator [OK]

Common Mistakes:

• Ignoring operator precedence
• Placing operators before operands
• Confusing postfix order

4. Given the following code snippet for infix to postfix conversion, what is the error?

if (ch == '+' || ch == '-' || ch == '*' || ch == '/') {
    while (!stack.isEmpty() && precedence(stack.peek()) >= precedence(ch)) {
        postfix += stack.pop();
    }
    stack.push(ch);
}

medium

A. The precedence comparison should use > instead of >=

B. The stack should be checked for null before peek()

C. Operands are not handled in this snippet

D. The while loop condition is incorrect for operator associativity

Solution

1. Step 1: Understand operator associativity in infix to postfix

   For left-associative operators like '-', '+', '*', '/', pop operators with precedence >= current operator.

2. Step 2: Identify issue with while condition

   Using >= causes right-associative operators like '^' to be handled incorrectly; condition should consider associativity.

3. Final Answer:

   The while loop condition is incorrect for operator associativity -> Option D

4. Quick Check:

   Associativity affects while condition [OK]

Hint: Check operator associativity when comparing precedence [OK]

Common Mistakes:

• Ignoring operator associativity
• Not handling operands separately
• Assuming peek() can be null without check

5. Convert the infix expression (A + B) * (C - D) / E to postfix using stack. What is the correct postfix expression?

hard

A. AB+CD*E-/

B. AB+CD-E*/

C. AB+CD-*E/

D. AB+CD-*E

Solution

1. Step 1: Convert each parenthesis group to postfix

   "(A + B)" becomes "AB+" and "(C - D)" becomes "CD-".

2. Step 2: Combine with multiplication and division respecting precedence

   Multiply results: "AB+CD-*" then divide by E: "AB+CD-*E/".

3. Final Answer:

   AB+CD-*E/ -> Option C

4. Quick Check:

   Parentheses handled first, then * and / [OK]

Hint: Convert inside parentheses first, then apply operators [OK]

Common Mistakes:

• Ignoring parentheses order
• Mixing operator precedence
• Dropping operators or operands