Question

How would you implement in Java the binary tree node class and the binary tree class to support the most efficient (from run-time perspective) equal check method (also has to be implemented):

    boolean equal(Node<T> root1, Node<T> root2) {}

or

    boolean equal(Tree t1, Tree t2) {}

First, I created the Node class as follows:

    public class Node<T> {
        private Node<T> left;
        private Node<T> right;
        private T data;

        // standard getters and setters
    }

and then the equals method that takes 2 root nodes as an arguments and runs the standard recursive comparison:

    public boolean equals(Node<T> root1, Node<T> root2) {
        boolean rootEqual = false;
        boolean lEqual = false;
        boolean rEqual = false;    

        if (root1 != null && root2 != null) {
            rootEqual = root1.getData().equals(root2.getData());

            if (root1.getLeft()!=null && root2.getLeft() != null) {
                // compare the left
                lEqual = equals(root1.getLeft(), root2.getLeft());
            }
            else if (root1.getLeft() == null && root2.getLeft() == null) {
                lEqual = true;
            }
            if (root1.getRight() != null && root2.getRight() != null) {
                // compare the right
                rEqual = equals(root1.getRight(), root2.getRight());
            }
            else if (root1.getRight() == null && root2.getRight() == null) {
                rEqual = true;
            }

            return (rootEqual && lEqual && rEqual);
        }
        return false;
    } 

My second attempt was to implement the trees using arrays and indexes for traversing. Then the comparison could be done using the bitwise operations (AND) on two arrays – read chunk from 2 arrays and mask one by another using logical AND. I failed to get my code working so I do not post it here (I’d appreciate your implementation of the second idea as well as your improvements).

Any thoughts how to do equality test for binary trees most efficiently?

EDIT

The question assumes structural equality. (Not the semantic equality)

However, code that tests the semantic equality e.g. “Should we consider the two trees to be equal if their contents are identical, even if their structure is not?” Would be just iterating over the tree in-order and it should be straightforward.

Answer 1

Well for one thing you’re always checking the branches, even if you spot that the roots are unequal. Your code would be simpler (IMO) and more efficient if you just returned false as soon as you spotted an inequality.

Another option to simplify things is to allow your equals method to accept null values and compare two nulls as being equal. That way you can avoid all those nullity checks in the different branches. This won’t make it more efficient, but it’ll be simpler:

public boolean equals(Node<T> root1, Node<T> root2) {
    // Shortcut for reference equality; also handles equals(null, null)
    if (root1 == root2) {
        return true;
    }
    if (root1 == null || root2 == null) {
        return false;
    }
    return root1.getData().equals(root2.getData()) &&
           equals(root1.getLeft(), root2.getLeft()) &&
           equals(root1.getRight(), root2.getRight());
} 

Note that currently this will fail if root1.getData() returns null. (That may or may not be possible with the way you’re adding nodes.)

EDIT: As discussed in comments, you could use hash codes to make a very quick “early out” – but it would add complexity.

Either you need to make your trees immutable (which is a whole other discussion) or you need each node to know about its parent, so that when the node is changed (e.g. by adding a leaf or changing the value) it needs to update its hash code and ask its parent to update too.