- Asked 3/14
- 2074 Bhadra · 10 marks
Convert A+B-C*(D-E+F/G)/H to postfix
Short answer
Scanning A+B-C*(D-E+F/G)/H with stack and precedence rules (pop higher/equal) gives postfix AB+CDE-FG/+*H/-.
Convert A + B - C * (D - E + F/G) / H expression into postfix expression using stack.
Rules: operand → output; ( → push; ) → pop to output until (; operator → pop operators of higher or equal precedence, then push; at the end pop everything. Precedence: * / above + -; equal precedence is evaluated left to right.
Infix: A + B - C * (D - E + F / G) / H
| Step | Symbol | Stack (bottom → top) | Postfix output |
|---|---|---|---|
| 1 | A | (empty) | A |
| 2 | + | + | A |
| 3 | B | + | AB |
| 4 | - | - | AB+ |
| 5 | C | - | AB+C |
| 6 | * | -* | AB+C |
| 7 | ( | -*( | AB+C |
| 8 | D | -*( | AB+CD |
| 9 | - | -*(- | AB+CD |
| 10 | E | -*(- | AB+CDE |
| 11 | + | -*(+ | AB+CDE- |
| 12 | F | -*(+ | AB+CDE-F |
| 13 | / | -*(+/ | AB+CDE-F |
| 14 | G | -*(+/ | AB+CDE-FG |
| 15 | ) | -* | AB+CDE-FG/+ |
| 16 | / | -/ | AB+CDE-FG/+* |
| 17 | H | -/ | AB+CDE-FG/+*H |
| end | (pop all) | (empty) | AB+CDE-FG/+*H/- |
Answer: AB+CDE-FG/+*H/-
Check with a tree: the expression is (A+B) − ((C * (…)) / H), where the bracket is (D−E) + (F/G). Post-order of this tree gives AB+CDE-FG/+*H/-.
✓ Answer checked by Claude Opus 5.5 (high effort)
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