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

Concept Flow - Infix to Postfix Conversion Using Stack

Read next token from infix

↓

Is token operand?

Yes→Add to postfix output

No↓

Is token '(' ?

Yes→Push '(' to stack

No↓

Is token ')' ?

Yes→Pop from stack to output until '('

No↓

Is token operator?

Yes→Pop operators from stack with higher/equal precedence to output

↓

Push current operator to stack

↓

Repeat until all tokens read

↓

Pop remaining operators from stack to output

↓

Postfix expression ready

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

Execution Sample

DSA C#

infix = "A+B*C"
stack = []
postfix = ""
for token in infix:
  if operand: postfix += token
  else if operator: pop higher prec from stack
  push token
pop all from stack to postfix

Converts infix expression A+B*C to postfix by scanning tokens, using stack for operators, and outputting operands immediately.

Execution Table

Step	Operation	Current Token	Stack Before	Stack After	Postfix Output	Visual State
1	Read token	A	[]	[]	A	Postfix: A | Stack: empty
2	Read token	+	[]	+	A	Postfix: A | Stack: +
3	Read token	B	+	+	AB	Postfix: AB | Stack: +
4	Read token	*	+	+, *	AB	Postfix: AB | Stack: + *
5	Read token	C	+, *	+, *	ABC	Postfix: ABC | Stack: + *
6	End of input		+, *		ABC*+	Pop * then + to postfix
7	Done				ABC*+	Final postfix expression

💡 All tokens processed and stack emptied to postfix output

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	Final
stack	[]	[]	+	+	+, *	+, *
postfix	""	A	A	AB	AB	ABC	ABC*+	ABC*+
current_token	null	A	+	B	*	C	end	end

Key Moments - 3 Insights

Why do we pop operators from the stack before pushing a new operator?

Because operators with higher or equal precedence must be output before the new operator to maintain correct order, as shown in step 4 and 6 where '*' is popped before pushing '+'

What happens when we encounter a parenthesis '(' or ')'?

When '(' is found, it is pushed to stack to mark a new sub-expression; when ')' is found, operators are popped until '(' is removed. This is not shown in this example but is critical for correct grouping.

Why are operands added directly to the postfix output?

Operands appear in the same order in postfix as in infix, so they are output immediately without stack, as seen in steps 1, 3, and 5.

Visual Quiz - 3 Questions

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

A ['*']

B ['+', '*']

C ['+']

D []

💡 Hint

Check row 4 in the execution_table for stack changes after reading '*'

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

A Step 5

B Step 3

C Step 4

D Step 6

💡 Hint

Look at the postfix output column in execution_table rows 3 to 6

If the input was 'A*B+C', how would the stack after reading '+' differ from the current example?

A Stack would be empty

B Stack would contain '+' and '*'

C Stack would contain only '+'

D Stack would contain only '*'

💡 Hint

Consider operator precedence and when operators are popped before pushing new ones

Concept Snapshot

Infix to Postfix Conversion Using Stack:
- Read tokens left to right
- Output operands immediately
- Push '(' to stack
- Pop until '(' on ')' token
- Pop higher/equal precedence operators before pushing new operator
- Pop all remaining operators at end
- Result is postfix expression

Full Transcript

This visualization shows how to convert an infix expression like A+B*C into postfix using a stack. Each token is read in order. Operands are added directly to the output. Operators are pushed to the stack, but before pushing, operators with higher or equal precedence are popped to maintain correct order. Parentheses control grouping by pushing '(' and popping until '(' on ')'. After all tokens are processed, remaining operators are popped to output. The example shows step-by-step stack and output changes, clarifying how the postfix expression ABC*+ is formed.

Practice

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

easy

A. To temporarily hold operators and manage their precedence

B. To store the final postfix expression

C. To reverse the infix expression

D. To count the number of operands

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: Differentiate stack usage from other tasks

   The postfix expression is built separately, and the stack is not used to store operands or reverse the expression.

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 [OK]

Common Mistakes:

• Thinking stack stores operands
• Confusing stack with output storage
• Assuming stack reverses expression

2. Which of the following C# code snippets correctly pushes an operator onto a stack named operatorStack?

easy

A. operatorStack.Push('+');

B. operatorStack.Add('+');

C. operatorStack.Insert('+');

D. operatorStack.Append('+');

Solution

1. Step 1: Recall stack method to add elements

   In C#, the method to add an element to a stack is Push().

2. Step 2: Verify other methods are invalid for Stack

   Add(), Insert(), and Append() are not valid methods for the Stack class.

3. Final Answer:

   operatorStack.Push('+'); -> Option A

4. Quick Check:

   Stack push method = Push() [OK]

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

Common Mistakes:

• Using Add() instead of Push()
• Confusing List methods with Stack methods
• Trying to use Append() on Stack

3. What is the postfix expression output for the infix expression A + B * C using standard operator precedence?

medium

A. *ABC+

B. AB+C*

C. ABC*+

D. A+BC*

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 respecting precedence

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

3. Final Answer:

   ABC*+ -> Option C

4. Quick Check:

   Multiplication before addition = ABC*+ [OK]

Hint: Multiply before add in postfix [OK]

Common Mistakes:

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

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

Stack<char> stack = new Stack<char>();
string infix = "A+B*C";
string postfix = "";
foreach (char ch in infix)
{
  if (char.IsLetter(ch))
    postfix += ch;
  else
    stack.Push(ch);
}
while (stack.Count > 0)
  postfix += stack.Pop();
Console.WriteLine(postfix);

medium

A. Operands are not added to postfix

B. Operators are pushed without checking precedence

C. Stack is not initialized

D. Postfix string is not printed

Solution

1. Step 1: Analyze operator handling in code

   The code pushes operators directly onto the stack without checking precedence or popping higher precedence operators first.

2. Step 2: Understand impact on postfix output

   This causes incorrect postfix order because operator precedence is ignored.

3. Final Answer:

   Operators are pushed without checking precedence -> Option B

4. Quick Check:

   Missing precedence check = B [OK]

Hint: Always check operator precedence before pushing [OK]

Common Mistakes:

• Ignoring operator precedence rules
• Assuming stack auto-sorts operators
• Not popping operators when needed

5. Convert the infix expression (A+B)*(C-D)/E to postfix using stack-based conversion. 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 stepwise to postfix

   A+B becomes AB+, C-D becomes CD-, then multiply: AB+CD-*, then divide by E: AB+CD-*E/.

3. Final Answer:

   AB+CD-*E/ -> Option D

4. Quick Check:

   Parentheses first, then multiply, then divide = AB+CD-*E/ [OK]

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

Common Mistakes:

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