Solving a Postfix Expression: ABCD+*/E- Tree

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SUMMARY

The discussion focuses on converting the postfix expression ABCD+*/E- into a binary tree structure. The key insight is that operands are pushed onto a stack, while operators pop the last two operands for computation. The operator '+' indicates that C and D become leaves of the '+' node, which is then a leaf of the '*' node. The root of the tree is determined to be '-' with '*' and '/E' as its children.

PREREQUISITES
  • Understanding of postfix notation and its evaluation
  • Familiarity with binary tree structures
  • Knowledge of stack data structures
  • Basic concepts of infix and postfix expressions
NEXT STEPS
  • Study the evaluation process of postfix expressions using stacks
  • Learn how to construct binary trees from postfix expressions
  • Explore the conversion between infix and postfix notations
  • Practice with additional examples of postfix expressions and their binary tree representations
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Students studying computer science, particularly those focusing on data structures and algorithms, as well as anyone needing to understand expression evaluation techniques.

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Homework Statement


I am having a hard time figuring out this postfix expression and turning it into a binary tree: ABCD+*/E-



The Attempt at a Solution



what's confusing me is the left side of the tree - the ABCD - I know the right side of the tree would look like this: - is the root and * / E

Any help would be greatly appreciated
 
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When you see a letter, think of it as an instruction to put it on a stack. When you see an operator, think of pulling the last two things off the stack and pushing the result of the operation back on the stack. So after you've parsed the ABCD, you have four things on the stack waiting for later operations.
 
What needs to be parsed? AB on the left and CD on the right?
 
Bucs44 said:
What needs to be parsed? AB on the left and CD on the right?

?? Neither. I can't start drawing a tree until I hit the +, at which point I realize C and D are leaves of the + node. Then the + node is itself a leaf of the * node, to get the other leaf I go back and see what's on the stack and there is a B etc etc.
 
(This belongs into the computer homework section.)

It might be instructive for you to work out the equivalent infix expression.
 

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