Postfix to Infix (original) (raw)

Last Updated : 27 Feb, 2025

Postfix to infix conversion involves transforming expressions where operators follow their operands (postfix notation) into standard mathematical expressions with operators placed between operands (infix notation). This conversion improves readability and understanding.

**Infix expression: The expression of the form a op b. When an operator is in-between every pair of operands.
**Postfix expression: The expression of the form a b op. When an operator is followed for every pair of operands.

**Examples:

**Input: abc++
**Output: (a + (b + c))
**Explanation: Infix expression is (a + (b + c)) for expression abc++

**Input: ab*c+
**Output: ((a*b)+c)
**Explanation: Infix expression is ((a*b)+c) for expression ab*c+

**Input: abc+*d/
**Output: (((a * (b + c))) / d)
**Explanation: Infix expression is (((a * (b + c)))/d) for expression abc+*d/

Try It Yourself

Approach: Using Stack - O(n) Time and O(n) Space

**Initialize Stack – Create an empty stack.

**Process Input Symbols – Read symbols one by one from the input

**Handle Operands – Push operands onto the stack.

**Handle Operators – Pop the top two values, form an infix expression, and push it back.

**Retrieve Final Expression – The remaining value in the stack is the desired infix expression.

C++ `

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

bool isOperand(char x) { return (x >= 'a' && x <= 'z') || (x >= 'A' && x <= 'Z'); }

// Get Infix for a given postfix // expression string getInfix(string exp) { stack s;

for (int i=0; exp[i]!='\0'; i++)
{
    // Push operands
    if (isOperand(exp[i]))
    {
       string op(1, exp[i]);
       s.push(op);
    }

    // We assume that input is
    // a valid postfix and expect
    // an operator.
    else
    {
        string op1 = s.top();
        s.pop();
        string op2 = s.top();
        s.pop();
        s.push("(" + op2 + exp[i] +
               op1 + ")");
    }
}

// There must be a single element
// in stack now which is the required
// infix.
return s.top();

}

// Driver code int main() { string exp = "ab*c+"; cout << getInfix(exp); return 0; }

Java

import java.util.*;

class GFG {

static boolean isOperand(char x) { return (x >= 'a' && x <= 'z') || (x >= 'A' && x <= 'Z'); }

// Get Infix for a given postfix // expression static String getInfix(String exp) { Stack s = new Stack();

for (int i = 0; i < exp.length(); i++)
{
    // Push operands
    if (isOperand(exp.charAt(i)))
    {
    s.push(exp.charAt(i) + "");
    }

    // We assume that input is
    // a valid postfix and expect
    // an operator.
    else
    {
        String op1 = s.peek();
        s.pop();
        String op2 = s.peek();
        s.pop();
        s.push("(" + op2 + exp.charAt(i) +
                op1 + ")");
    }
}

// There must be a single element
// in stack now which is the required
// infix.
return s.peek();

}

// Driver code public static void main(String args[]) { String exp = "ab*c+"; System.out.println( getInfix(exp)); } }

// This code is contributed by Arnab Kundu

Python

def isOperand(x): return ((x >= 'a' and x <= 'z') or (x >= 'A' and x <= 'Z'))

Get Infix for a given postfix

expression

def getInfix(exp) :

s = [] 

for i in exp:     
    
    # Push operands 
    if (isOperand(i)) :         
        s.insert(0, i) 
        
    # We assume that input is a 
    # valid postfix and expect 
    # an operator. 
    else:
    
        op1 = s[0] 
        s.pop(0) 
        op2 = s[0] 
        s.pop(0) 
        s.insert(0, "(" + op2 + i +
                         op1 + ")") 
        
# There must be a single element in 
# stack now which is the required 
# infix. 
return s[0]

Driver Code

if name == 'main':

exp = "ab*c+"
print(getInfix(exp.strip()))

This code is contributed by

Shubham Singh(SHUBHAMSINGH10)

C#

using System; using System.Collections;

class GFG {

static Boolean isOperand(char x) { return (x >= 'a' && x <= 'z') || (x >= 'A' && x <= 'Z'); }

// Get Infix for a given postfix // expression static String getInfix(String exp) { Stack s = new Stack();

for (int i = 0; i < exp.Length; i++)
{
    // Push operands
    if (isOperand(exp[i]))
    {
        s.Push(exp[i] + "");
    }

    // We assume that input is
    // a valid postfix and expect
    // an operator.
    else
    {
        String op1 = (String) s.Peek();
        s.Pop();
        String op2 = (String) s.Peek();
        s.Pop();
        s.Push("(" + op2 + exp[i] +
                op1 + ")");
    }
}

// There must be a single element
// in stack now which is the required
// infix.
return (String)s.Peek();

}

// Driver code public static void Main(String []args) { String exp = "ab*c+"; Console.WriteLine( getInfix(exp)); } }

// This code is contributed by Arnab Kundu

JavaScript

function isOperand(x) { return (x >= 'a' && x <= 'z') || (x >= 'A' && x <= 'Z'); }

// Get Infix for a given postfix // expression function getInfix(exp) { let s = [];

for (let i = 0; i < exp.length; i++)
{
    // Push operands
    if (isOperand(exp[i]))
    {
    s.push(exp[i] + "");
    }

    // We assume that input is
    // a valid postfix and expect
    // an operator.
    else
    {
        let op1 = s[s.length-1];
        s.pop();
        let op2 = s[s.length-1];
        s.pop();
        s.push("(" + op2 + exp[i] +
                op1 + ")");
    }
}

// There must be a single element
// in stack now which is the required
// infix.
return s[s.length-1];

}

// Driver code let exp = "ab*c+"; console.log( getInfix(exp));