In this tutorial We have explored an algorithm to convert a given Postfix expression to Infix expression using Stack.

## Algorithm For Postfix to Infix Conversion

Iterate the given expression from left to right, one character at a time Step 1 : If a character is operand, push it to stack. Step 2: If a character is an operator, if there are fewer than 2 values on the stack give error "insufficient values in expression" goto Step 4 else pop 2 operands from stack create a new string and by putting the operator between operands. push this string into stack Repeat Steps 1 and 2 Step 3: At last there will be only one value or one string in the stack which will be our infix expression Step 4: Exit

## Some important terminology

**Postfix Expression**

In Postfix Expression operator appear after operand, this expression is known as Postfix operation.**Infix**

If Infix Expression operator is in between of operands, this expression is known as Infix operation.

## Steps to Convert Postfix to Infix

- Start Iterating the given Postfix Expression from Left to right
- If Character is operand then push it into the stack.
- If Character is operator then pop top 2 Characters which is operands from the stack.
- After poping create a string in which comming operator will be in between the operands.
- push this newly created string into stack.
- Above process will continue till expression have characters left
- At the end only one value will remain if there is integers in expressions. If there is character then one string will be in output as infix expression.

## Example to convert postfix to Infix

**Postfix Expression :** abc-+de-+

Token | Stack | Action |

a | a | push a in stack |

b | a, b | push b in stack |

c | a, b, c | push c in stack |

– | a , b – c | pop b and c from stack and put – in between and push into stack |

+ | a + b – c | pop a and b-c from stack and put + in between and push into stack |

d | a + b – c, d | push d in stack |

e | a + b – c, d , e | push e in stack |

– | a + b – c, d – e | pop d and e from stack and put – in between and push into stack |

+ | a + b – c + d – e | pop a + b – c and d – e from stack and put + in between and push into stack |

## Solution for postfix expression

**postfix expression:** 752+*415-/-

Token | Stack | Action |

7 | 7 | push 7 in stack |

5 | 7, 5 | push 5 in stack |

2 | 7 , 5, 2 | push 2 in stack |

+ | 7, 7 | pop 2 and 5 from stack, sum it and then again push it |

* | 49 | pop 7 and 7 from stack and multiply it and then push it again |

4 | 49, 4 | push 4 in stack |

1 | 49, 4, 1 | push 1 in stack |

5 | 49, 4, 1, 5 | push 5 in stack |

– | 49, 4, -4 | pop 5 and 1 from stack |

/ | 49, -1 | pop 4 and 4 from stack |

– | 50 | pop 1 and 49 from stack |

## Infix to postfix conversion program

import java.util.*; public class Main { public static String convert(String exp){ int len = exp.length(); Stack<String> stack = new Stack<>(); for (int i = 0; i <len ; i++) { char c = exp.charAt(i); if(c=='*'||c=='/'||c=='^'||c=='+'||c=='-' ){ String s1 = stack.pop(); String s2 = stack.pop(); String temp = "("+s2+c+s1+")"; stack.push(temp); }else{ stack.push(c+""); } } String result=stack.pop(); return result; } public static void main(String[] args) { Scanner sc = new Scanner(System.in); System.out.println("Please enter Postfix Expression: "); String exp = sc.nextLine(); System.out.println("Infix Expression: " + Main.convert(exp)); } }