use a binary tree to encode infix arithmetic expressions on integers

Question

This is a question from aws educate. I have been thinking about this for a long time and I am not really getting anywhere.

You want to use a binary tree to encode infix arithmetic expressions on integers. Operations are addition and multiplication Draw a picture of what the tree looks like. Write a class definition. Write an evaluate() member function. How would you make your evaluate() iterative instead of recursive

If I could get an explanation that would be fine or some example too

score 1 · Answer 1 · answered Sep 21 '17 at 23:34

The question is asking you to write a tree class which can represent expressions like "2 + 2" or "3 * 1 + 5". So the class represents a tree which has a root and internal nodes which correspond with applications of the "*" or "+" operators. The leaf nodes will correspond to integer values like "5" or "2" which are being operated upon. A typical evaluation function which would produce a result from such a tree might be recursive. They're asking you to also consider how you can arrive at a result iteratively. Such an iterative approach might involve adding nodes successively to a queue or stack data structure and popping them off one-by-one to be handled somehow.

See here: https://en.wikipedia.org/wiki/Tree_traversal

score 1 · Accepted Answer · answered May 26 '18 at 21:00

Illustration - binary tree to encode infix arithmetic expressions on integers

As you can see, the leafs are the value (or literal) and the other nodes contain arithmetic operators (+, - , /, *)

Some Code - recusion

In Java, you could use the recursion to solve this problem (under the condition that the tree's height is not to big)

public class Main {

    public static void main(String[] args) {
        // op1=(1 + 2)
        Node op1 = new Node(1, "+", 2);
        System.out.println("op1="+op1.evaluate()); // op1=3

        // op2=(1 + 2) + 3
        Node op2 = new Node(op1, "+", 3);
        System.out.println("op2="+op2.evaluate()); // op2=6

        // op3=(4 * 5)
        Node op3 = new Node(4, "*", 5);
        System.out.println("op3="+op3.evaluate()); // op3=20

        // op4=((1+2)+3)*(4*5)
        Node op4 = new Node(op2, "*", op3);
        System.out.println("op4="+op4.evaluate()); // op4=120
    }
}

class Node {

    private Node left;
    private Node right;
    private String operatorOrLiteral;

    public Node(String value){
        this.operatorOrLiteral = value;
    }

    public Node(Node left, String operation, Node right){
        this.operatorOrLiteral = operation;
        this.left = left;
        this.right = right;
    }

    public Node(int literal1, String operation, int literal2){
        this(new Node(Integer.toString(literal1)), operation, new Node(Integer.toString(literal2)));
    }

    public Node(Node left, String operation, int literal2) {
        this(left, operation, new Node(Integer.toString(literal2)));
    }

    public Node(int literal1, String operation, Node right) {
        this(new Node(Integer.toString(literal1)), operation, right);
    }

    public int evaluate(){
        if(isLiteral()) {
            return Integer.parseInt(operatorOrLiteral);
        }

        switch (operatorOrLiteral) {
            case "*": return left.evaluate() * right.evaluate();
            case "/": return left.evaluate() / right.evaluate();
            case "-": return left.evaluate() - right.evaluate();
            case "+": return left.evaluate() + right.evaluate();
            default: throw new IllegalArgumentException(operatorOrLiteral + " is not recognised");
        }
    }

    private boolean isLiteral() {
        return left == null && right == null;
    }

    public String toString() {
        if(isLiteral()) {
            return operatorOrLiteral;
        }

        return "(" + left.toString() + operatorOrLiteral + right.toString() + ")";
    }
}

Some Code - iterative

Or as @David Sanders as mentioned, you can work with tree traversal.

use a binary tree to encode infix arithmetic expressions on integers

2 Answers2