Complete the following methods:

nodeExists(int idx) -- this should return true if the tree has an element at index idx and false when there is no such element

leftChild(int idx) -- returns the element that is the left child of the one at index idx. Your method should return null if idx does not have a left child.

parent(int idx) -- returns the element that is the parent of the one at index idx. Your method should return null if idx does not have a parent.

 package edu.buffalo.cse116; import java.util.AbstractSet; import java.util.Iterator; /** * Implementation of an abstract set using an array-based binary tree. This is used to help teach binary trees and will * have more details explained in future lectures. * * @author William J. Collins * @author Matthew Hertz * @param  Data type (which must be Comparable) of the elements in this tree. */ public class ArrayBinaryTree> extends AbstractSet { /** Index where the root node can be found. */ private static final int ROOT = 0; /** Array used to store the elements in the binary tree. */ private E[] store; /** Number of elements within the tree. */ private int size; /** * Initializes this ArrayBinaryTree object to be empty. This creates the array in which items will be stored. */ @SuppressWarnings("unchecked") public ArrayBinaryTree() { store = (E[]) new Comparable[31]; size = 0; } /** * Checks if the binary tree contains an element at the given index. This requires checking both that the array is * large enough (to avoid triggering an exception) AND (when the array is large enough) that the array has a non-null * value at that index. * * @param idx Index to be checked out. * @return True if there is an element at the given index; false otherwise. */ private boolean nodeExists(int idx) { } /** * Given an index, returns the element in that node's left child. If the left child node does not exist, null should * be returned. It is important that this NOT trigger an index out of bounds exception. * * @param idx Index of the node for which we want the left child. * @return Value of the node's left child or null if no left child exists. */ private E leftChild(int idx) { } /** * Given an index, returns the value of that node's parent. If the node is the root (and so has no parent), null * should be returned. It is important that this NOT trigger an index out of bounds exception. * * @param idx Index of the node for which we want the parent. * @return Value of the node's parent or null if no parent exists. */ private E parent(int idx) { } /** * Returns the size of this ArrayBinaryTree object. * * @return the size of this ArrayBinaryTree object. */ @Override public int size() { return size; } /** * Returns an iterator that will return the elements in this ArrayBinaryTree, but without any specific ordering. * * @return Iterator positioned at the smallest element in this ArrayBinaryTree object. */ @Override public Iterator iterator() { // Skipped for now. throw new UnsupportedOperationException(); } /** * Determines if there is at least one element in this ArrayBinaryTree object that equals a specified element. * * @param obj - the element sought in this ArrayBinaryTree object. * @return true - if there is an element in this ArrayBinaryTree object that equals obj; otherwise, return false. * @throws ClassCastException - if obj cannot be compared to the elements in this ArrayBinaryTree object. * @throws NullPointerException - if obj is null. */ @Override public boolean contains(Object obj) { int index = getIndex(obj); return index != -1; } /** * Finds the index at which a specified element exist or -1 if the object is not in the tree. * * @param obj Element whose Entry is sought. * @return Element in the treeEntry object in the ArrayBinaryTree that holds obj; if no Entry exists, returns null. */ protected int getIndex(Object obj) { // Handle the special case of obj being null -- we can return null since we // KNOW this not to be in the tree. By definition, we are to throw this exception. if (obj == null) { throw new NullPointerException(); } // Keep searching while there might be a node at the current index within the tree for (int idx = ROOT; idx < store.length; idx++ ) { // If we have fallen off the bottom of the tree, obj cannot be found // within it. if (store[idx] == null) { return -1; } E curElement = store[idx]; @SuppressWarnings("unchecked") int comp = curElement.compareTo((E) obj); if (comp == 0) { return idx; } } // We made it through the tree and did not find the element; return that it was not found. return -1; } }