For Postfix to Infix, expression conversion push each character one by one into the stack. whenever operator occurs pop the top 2 elements from the stack and again push back to the stack till last.
steps are as follows:
Expression |
Stack |
$ABC+-D*$ |
NULL |
$BC+-D*$ |
$A$ |
$C+-D*$ |
$A, B$ |
$+-D*$ |
$A,B,C$ |
$-D*$ |
$A,(B+C),$ |
$D*$ |
$(A-(B+C)),$ |
$*$ |
$(A-(B+C)),D$ |
$$ |
$(A-(B+C))*D$ |
So option A is correct here.