# Convert Infix to Prefix Expression

To solve expressions by the computer, we can either convert it in postfix form or to the prefix form. Here we will see how infix expressions are converted to prefix form.

At first infix expression is reversed. Note that for reversing the opening and closing parenthesis will also be reversed.

for an example: The expression: A + B * (C - D)

after reversing the expression will be: ) D – C ( * B + A

so we need to convert opening parenthesis to closing parenthesis and vice versa.

After reversing, the expression is converted to postfix form by using infix to postfix algorithm. After that again the postfix expression is reversed to get the prefix expression.

## Input and Output

Input:
Infix Expression: x^y/(5*z)+2
Output:
Prefix Form Is: +/^xy*5z2

## Algorithm

infixToPrefix(infix)

Input − Infix expression to convert into prefix form.

Output − The prefix expression.

Begin
reverse the infix expression
for each character ch of reversed infix expression, do
if ch = opening parenthesis, then
convert ch to closing parenthesis
else if ch = closing parenthesis, then
convert ch to opening parenthesis
done

postfix := find transformed infix expression to postfix expression
prefix := reverse recently calculated postfix form
return prefix
End

## Example

#include<iostream>
#include<stack>
#include<locale> //for function isalnum()
#include<algorithm>
using namespace std;

int preced(char ch) {
if(ch == '+' || ch == '-') {
return 1;    //Precedence of + or - is 1
}else if(ch == '*' || ch == '/') {
return 2;    //Precedence of * or / is 2
}else if(ch == '^') {
return 3;    //Precedence of ^ is 3
}else {
return 0;
}
}

string inToPost(string infix) {
stack<char> stk;
stk.push('#');    //add some extra character to avoid underflow
string postfix = "";   //initially the postfix string is empty
string::iterator it;

for(it = infix.begin(); it!=infix.end(); it++) {
if(isalnum(char(*it)))
postfix += *it;    //add to postfix when character is letter or number
else if(*it == '(')
stk.push('(');
else if(*it == '^')
stk.push('^');
else if(*it == ')') {
while(stk.top() != '#' && stk.top() != '(') {
postfix += stk.top();    //store and pop until ( has found
stk.pop();
}

stk.pop();    //remove the '(' from stack
}else {
if(preced(*it) > preced(stk.top()))
stk.push(*it);    //push if precedence is high
else {
while(stk.top() != '#' && preced(*it) <= preced(stk.top())) {
postfix += stk.top();    //store and pop until higher precedence is found
stk.pop();
}
stk.push(*it);
}
}
}

while(stk.top() != '#') {
postfix += stk.top();    //store and pop until stack is not empty

stk.pop();

}
return postfix;
}

string inToPre(string infix) {
string prefix;
reverse(infix.begin(), infix.end());    //reverse the infix expression
string::iterator it;

for(it = infix.begin(); it != infix.end(); it++) {    //reverse the parenthesis after reverse
if(*it == '(')
*it = ')';
else if(*it == ')')
*it = '(';
}

prefix = inToPost(infix);                 //convert new reversed infix to postfix form.
reverse(prefix.begin(), prefix.end());    //again reverse the result to get final prefix form
return prefix;
}

int main() {
string infix = "x^y/(5*z)+2";
cout << "Prefix Form Is: " << inToPre(infix) << endl;
}

## Output

Prefix Form Is: +/^xy*5z2