Infix to Postfix Conversion Using Stack in DSA C++ - Execution Trace

Concept Flow - Infix to Postfix Conversion Using Stack

Start with empty stack and output

↓

Read next symbol from infix

↓

Is symbol operand?

Yes→Add to output

No↓

Is symbol '('?

Yes→Push '(' on stack

No↓

Is symbol ')' ?

Yes→Pop and output until '(' found

No↓

Is symbol operator?

↓

While stack top has higher or equal precedence

↓

Pop from stack to output

↓

Push current operator on stack

↓

Repeat until infix ends

↓

Pop all remaining stack to output

↓

End with postfix expression

The flow reads each symbol from the infix expression, decides if it is operand, operator, or parenthesis, and uses a stack to reorder operators by precedence to produce postfix.

Execution Sample

DSA C++

infix = "A+B*C"
stack = []
output = ""
for symbol in infix:
  if operand: output += symbol
  else if operator: handle stack
pop remaining stack to output

Converts infix expression A+B*C to postfix by scanning each symbol and using stack to reorder operators.

Execution Table

Step	Operation	Symbol	Stack State	Output State	Visual State
1	Read symbol	A	[]	A	Output: A; Stack: empty
2	Read symbol	+	[+]	A	Push '+' on stack; Output unchanged
3	Read symbol	B	[+]	AB	Output: AB; Stack: '+'
4	Read symbol	*	[+, *]	AB	Push '*' on stack; Output unchanged
5	Read symbol	C	[+, *]	ABC	Output: ABC; Stack: '+', '*'
6	End of input		[+, *]	ABC	Pop '*' to output: ABC*; Stack: '+'
7	Pop remaining		[+]	ABC*	Pop '+' to output: ABC*+; Stack empty
8	End		[]	ABC*+	Final postfix expression: ABC*+

💡 All symbols processed and stack emptied to output

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	Final
symbol		A	+	B	*	C
stack	[]	[]	[+]	[+]	[+, *]	[+, *]	[+]	[]	[]
output	""	A	A	AB	AB	ABC	ABC*	ABC*+	ABC*+

Key Moments - 3 Insights

Why do we push '(' directly on the stack without popping?

Because '(' marks a new sub-expression boundary; we push it to know when to stop popping operators when ')' is found, as shown in the concept flow and would appear in stack states if parentheses were present.

Why do we pop operators from the stack when the incoming operator has lower or equal precedence?

To maintain correct operator order in postfix, we pop higher or equal precedence operators first before pushing the new one, as seen in step 6 where '*' is popped before '+' to output.

Why do operands go directly to output without stack involvement?

Operands appear in postfix in the same order as infix, so we add them directly to output immediately, as shown in steps 1, 3, and 5.

Visual Quiz - 3 Questions

Look at the execution table, what is the stack state after reading symbol '*'?

A [+, *]

B [+]

C []

D [*]

💡 Hint

Check step 4 in the execution_table under 'Stack State'

At which step does the output first include the '*' operator?

A Step 5

B Step 6

C Step 7

D Step 4

💡 Hint

Look at 'Output State' column in execution_table rows 5, 6, and 7

If the input was 'A*B+C', how would the stack state change after reading '+' compared to the current example?

A Stack would pop '*' before pushing '+'

B Stack would push '+' without popping

C Stack would be empty after reading '+'

D Stack would contain only '*' and '+' together

💡 Hint

Recall operator precedence and popping rules from concept_flow and key_moments

Concept Snapshot

Infix to Postfix Conversion Using Stack:
- Scan infix left to right
- Output operands immediately
- Push '(' on stack
- Pop until '(' on ')' encountered
- For operators, pop higher/equal precedence first
- Push current operator
- After input ends, pop all stack to output
- Result is postfix expression

Full Transcript

This visualization shows how to convert an infix expression like A+B*C into postfix using a stack. We read each symbol: operands go directly to output, operators are pushed on stack but first pop any operators with higher or equal precedence. Parentheses are handled by pushing '(' and popping until '(' when ')' is found. After reading all symbols, we pop remaining operators to output. The execution table tracks each step, stack state, and output. Key moments clarify why operands skip stack, why operators pop based on precedence, and how parentheses control popping. The quiz tests understanding of stack states and output changes during conversion.

Practice

1. What is the main purpose of using a stack in Infix to Postfix Conversion?

easy

A. To count the number of operators

B. To temporarily hold operators and manage their precedence

C. To reverse the order of the expression

D. To store operands permanently

Solution

1. Step 1: Understand the role of stack in conversion

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

2. Step 2: Identify what is stored in the stack

   Operands are directly added to the output, while operators are pushed to the stack to manage order and precedence.

3. Final Answer:

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

4. Quick Check:

   Stack holds operators = D [OK]

Hint: Stack holds operators to manage precedence [OK]

Common Mistakes:

• Thinking stack stores operands
• Believing stack reverses the entire expression
• Confusing stack role with counting operators

2. Which of the following is the correct syntax to push an operator onto a stack named st in C++?

easy

A. st.push(operator);

B. push(st, operator);

C. st.add(operator);

D. push.operator(st);

Solution

1. Step 1: Recall C++ stack push syntax

   In C++, the standard stack uses the member function push() to add elements.

2. Step 2: Match correct syntax

   The correct syntax is st.push(operator); where st is the stack object.

3. Final Answer:

   st.push(operator); -> Option A

4. Quick Check:

   Stack push method = st.push() [OK]

Hint: Use st.push(element) to add to stack [OK]

Common Mistakes:

• Using non-member functions like push(st, element)
• Using incorrect method names like add()
• Incorrect dot notation like push.operator()

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

medium

A. A+BC*

B. AB+C*

C. ABC*+

D. *ABC+

Solution

1. Step 1: Apply operator precedence rules

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

2. Step 2: Convert to postfix

   Postfix places operands first, then operator: B C * then A +, resulting in ABC*+.

3. Final Answer:

   ABC*+ -> Option C

4. Quick Check:

   Postfix of A + B * C = ABC*+ [OK]

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

Common Mistakes:

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

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

std::stack<char> st;
std::string infix = "A+B*C";
std::string postfix = "";
for (char ch : infix) {
  if (isalnum(ch))
    postfix += ch;
  else {
    while (!st.empty() && st.top() >= ch) {
      postfix += st.top();
      st.pop();
    }
    st.push(ch);
  }
}
while (!st.empty()) {
  postfix += st.top();
  st.pop();
}

medium

A. Operator precedence comparison is incorrect

B. Operands are not added to postfix

C. Stack is not used for operators

D. Postfix string is not initialized

Solution

1. Step 1: Analyze operator precedence check

   The code compares operators using >= directly on characters, which does not correctly reflect operator precedence.

2. Step 2: Identify correct precedence handling

   Operator precedence should be checked using a function that assigns precedence values, not character comparison.

3. Final Answer:

   Operator precedence comparison is incorrect -> Option A

4. Quick Check:

   Operator precedence needs function, not char compare [OK]

Hint: Use precedence function, not char comparison [OK]

Common Mistakes:

• Comparing operators as characters
• Ignoring precedence rules
• Not popping stack correctly

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: Handle parentheses and operator precedence

   Parentheses force A+B and C-D to be evaluated first, then multiply results, then divide by E.

2. Step 2: Convert step-by-step to postfix

   A B + (for A+B), C D - (for C-D), then multiply: AB+CD-*, finally divide by E: AB+CD-*E/.

3. Final Answer:

   AB+CD-*E/ -> Option D

4. Quick Check:

   Postfix of (A+B)*(C-D)/E = AB+CD-*E/ [OK]

Hint: Process parentheses first, then operators by precedence [OK]

Common Mistakes:

• Ignoring parentheses order
• Mixing operator order
• Placing division before multiplication