Postfix to Prefix Conversion Program using Stack in C++

July 30, 2025

Postfix to Prefix Conversion in C++

Postfix to Prefix Conversion in C++ – Prefix, Postfix and Infix are the different ways to write expressions as notations. In infix, they are normal notations as used by mathematical expressions in copies. In Prefix expression, the  operator is prefixed to operands, and in Postfix, or Reverse Polish Notation, the operator comes after the operands. In this article we will know how to perform Post fix expressions to prefix expressions converstion using a stack in C++.

Postfix to prefix expression

What is Postfix to Prefix Conversion in C++?

Infix: (X + Y)

  • Postfix – The postfix will look like, XY+
  • Prefix: The prefix will look like, +YX

Infix : (X + Y) / (U – V)

  • Postfix – The postfix will look like, XY+UV-/
  • Prefix – The prefix will look like, /+XY-UV

Here we need to use a stack data structure. SInce we have our own inbuilt stack DS in C++ STL, we are going to use it. If you use want to know more about that : Stack in C++ STL 

Follow this algorithm to solve this problem

There may be  many ways to find the post fix to prefix, here it goes one easy way to imlement an algorithm to do so.

  • Read the expression from left to right.
  • If there is any operand present, push it to a stack.
  • Then if you come accross an operator, pop two of the opearands from the stack and concatinate them like,
    the operator + 2nd top value of the stack + first top value of the stack.
  • Repeat the process.
  • Reverse the string at the end.

Fun fact: This can be interpreted as a binary tree if you can make a binary tree with operators and add the operands at last.

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Stack Representation in C++

Representation of a Stack as an Array

Representation of a Stack as a Linked List

Infix to postfix conversion

Infix to prefix expression

Visual Implementation of the Algorithm :

Suppose the expression in Infix is (X+Y)/(U-V), and in Postfix: XY+UV-/ 

Here all the iterations are shown below.

Postfix to prefix conversion-2
Postfix to prefix conversion – 1

The code to implement this:

Run

#include <bits/stdc++.h>
using namespace std;

bool isOperator(char x) {
    return (x == '+' || x == '-' || x == '*' || x == '/');
}

string postToPre(string post_exp) {
    stack s;

    for (int i = 0; i < post_exp.length(); i++) {
        char ch = post_exp[i];

        if (isOperator(ch)) {
            string op2 = s.top(); s.pop();
            string op1 = s.top(); s.pop();
            string temp = ch + op1 + op2;
            s.push(temp);
        } else {
            s.push(string(1, ch));
        }
    }

    return s.top();
}

int main() {
    string postfix;
    cout << "Enter the Postfix Expression: "; cin >> postfix;

    string prefix = postToPre(postfix);
    cout << "The Prefix expression is: " << prefix << endl;

    return 0;
}

Input :

ABC*+

Output:

+ A * B C

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FAQs - Postfix to Prefix Conversion in C++

FAQs - Postfix to Prefix Conversion in C++

In Postfix to Prefix conversion, each subexpression (like +AB) can be more than one character. Using a stack<string> allows handling full subexpressions and prevents incorrect concatenation or type errors.

Prefix notation inherently preserves operator precedence by placing the operator before its operands. The conversion process constructs expressions recursively using the stack, thus eliminating the need for explicit parentheses.

Edge cases include: insufficient operands, extra operators, invalid characters, or single-character input. These should be validated before processing, and the stack should always contain exactly one element at the end for a valid expression.

The algorithm naturally handles nested sub-expressions by using a stack: when an operator is encountered, the top two operands (which can be full sub-expressions) are popped, combined, and pushed back—thus preserving correct nesting order.

Postfix expressions are evaluated from left to right because operands appear before the operator. This order ensures the top two elements on the stack are always the correct operands for the current operator being processed.

Both time and space complexity are O(n), where n is the length of the postfix expression. The size of intermediate expressions and the number of operators affect the space required by the stack during execution.

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Priority Queue

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