The following algorithm transforms the infix expression X into its equivalent postfix expression Y. The algorithm uses a stack to temporarily hold operators and left parentheses. The postfix expression Y will be constructed from left to right using the operands from X and the operators which are removed from STACK.

We begin by pushing a left parenthesis onto STACK and adding a right parenthesis at the end of X. The algorithm is completed when STACK is empty.

**Algorithm**

```
Suppose X is an arithmetic expression written in infix notation. This algorithm finds the equivalent postfix expression Y.
1. push "(" onto stack, and add ")" to the end of X.
2. scan X from left to right and repeat Steps 3 to 6 for each element of X until the STACK is empty :
3. if an operand is encountered, add it to Y.
4. if a left parenthesis is encountered, push it onto STACK.
5. if an operator is encountered, then
(a) Repeatedly pop from STACK and add to Y each operator (on the top of STACK) which has the same precedence as or higher precedence than operator.
(b) Add operator to STACK.
/*End of If structure */
6. if a right parenthesis is encountered, then :
(a) Repeatedly pop from STACK and add to Y each operator (on the top of STACK) until a left parenthesis is encountered.
(b) Remove the left parenthesis. [Do not add the left parenthesis to Y].
/* end of If structure */
/* end of Step 2 loop */
7. exit.
```

This is an Algorithm to Convert Infix expression to Postfix. If you have any questions then comment below.

conversion from prefix to postfix evaluation.

1.(A+B)*(D/E)

2.(A+B^D)/E-F/G