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Home/ Questions/Q 8647523
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T13:05:43+00:00 2026-06-12T13:05:43+00:00

im trying to create a method which allows me to check if a BST

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im trying to create a method which allows me to check if a BST contains an item. this is what i have so far:

public boolean contains(Object item) {
    // YOUR CODE HERE

    //changes item to E so it can be used in the comparator
    E value1 = (E) item;

    if (root.value.equals(item)){
        return true;
    }

    comp.compare(value1,root.value);
    if(value1<root.value){
        if (left == null)
            return null;
        else
            return left.contains(item);
    }

    else if(item >= value){
        if (right == null)
            return null;
        else
            return right.contains(item);
    }
}

and these are my fields:

// Data fields
private BSTNode root;
private int count = 0;
private Comparator<E> comp;   // default comparator

/** Private class for the nodes.
 *  Has public fields so methods in BSTSet can access fields directly. 
 */
private class BSTNode {

    // Data fields

    public E value;
    public BSTNode left = null;
    public BSTNode right = null;

    // Constructor

    public BSTNode(E v) {
        value = v;
    }

}


public BSTSet() {
    comp = new ComparableComparator();      // Declared below
}

public BSTSet(Comparator <E> c) {
    comp = c;
}

my question is how do i fix my contains method so that it works. So far it gets to the line under comp.compare(value1.root.value)) and says that ‘<‘ cannot be used on two elements of type E. how do i fix this so that i can continue to run the comparator?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-12T13:05:44+00:00Added an answer on June 12, 2026 at 1:05 pm

    Your comparator returns an int.

    If this int is 0, than the value of both objects is the same, if it’s less than 0, it’s equivalente to your <, if it’s more than 0, it’s equivalent to your >.

    You need to use the result of your comparator call to check if you should traverse the right or left sub-tree.

    Also, please don’t return null if the left/right branch is null, return false. That’s the correct semantics of the algorithm, if there is no left branch and the value is smaller than the current root, than the item is not in the tree, hence, false.

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