To evaluate an expression without any embedded function calls:
(A) One stack is enough
(B) Two stacks are needed
(C) As many stacks as the height of the expression tree are needed
(D) A Turing machine is needed in the general case
Any expression can be converted into Postfix or Prefix form.
Prefix and postfix evaluation can be done using a single stack.
For example : Expression ’10 2 8 * + 3 -‘ is given.
PUSH 10 in the stack.
PUSH 2 in the stack.
PUSH 8 in the stack.
When operator ‘*’ occurs, POP 2 and 8 from the stack.
PUSH 2 * 8 = 16 in the stack.
When operator ‘+’ occurs, POP 16 and 10 from the stack.
PUSH 10 * 16 = 26 in the stack.
PUSH 3 in the stack.
When operator ‘-‘ occurs, POP 26 and 3 from the stack.
PUSH 26 – 3 = 23 in the stack.
So, 23 is the answer obtained using single stack.
Thus, option (A) is correct.
Please comment below if you find anything wrong in the above post.
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