Very confused on this Java code would be great thanks Parent

Very confused on this... Java code would be great, thanks!

(Parent reference for BST) Redefine TreeNode by adding a reference to a
node’s parent, as shown below:

BST.TreeNode
#element: E
#left: TreeNode
#right: TreeNode
#parent: TreeNode

Reimplement the insert and delete methods in the BST class to update the
parent for each node in the tree. Add the following new method in BST:

/** Returns the node for the specified element.
* Returns null if the element is not in the tree. */
private TreeNode getNode(E element)

/** Returns true if the node for the element is a leaf */
private boolean isLeaf(E element)

/** Returns the path of elements from the specified element
* to the root in an array list. */
public ArrayList getPath(E e)

Write a test program that prompts the user to enter 10 integers, adds them to
the tree, deletes the first integer from the tree, and displays the paths for all leaf
nodes. Here is a sample run:
Enter 10 integers: 45 54 67 56 50 45 23 59 23 67
[50, 54, 23]
[59, 56, 67, 54, 23]

Solution

public class BinarySearchTree { public static Node root; public BinarySearchTree(){ this.root = null; } public boolean find(int id){ Node current = root; while(current!=null){ if(current.data==id){ return true; }else if(current.data>id){ current = current.left; }else{ current = current.right; } } return false; } public boolean delete(int id){ Node parent = root; Node current = root; boolean isLeftChild = false; while(current.data!=id){ parent = current; if(current.data>id){ isLeftChild = true; current = current.left; }else{ isLeftChild = false; current = current.right; } if(current ==null){ return false; } } //if i am here that means we have found the node //Case 1: if node to be deleted has no children if(current.left==null && current.right==null){ if(current==root){ root = null; } if(isLeftChild ==true){ parent.left = null; }else{ parent.right = null; } } //Case 2 : if node to be deleted has only one child else if(current.right==null){ if(current==root){ root = current.left; }else if(isLeftChild){ parent.left = current.left; }else{ parent.right = current.left; } } else if(current.left==null){ if(current==root){ root = current.right; }else if(isLeftChild){ parent.left = current.right; }else{ parent.right = current.right; } }else if(current.left!=null && current.right!=null){ //now we have found the minimum element in the right sub tree Node successor = getSuccessor(current); if(current==root){ root = successor; }else if(isLeftChild){ parent.left = successor; }else{ parent.right = successor; } successor.left = current.left; } return true; } public Node getSuccessor(Node deleleNode){ Node successsor =null; Node successsorParent =null; Node current = deleleNode.right; while(current!=null){ successsorParent = successsor; successsor = current; current = current.left; } //check if successor has the right child, it cannot have left child for sure // if it does have the right child, add it to the left of successorParent. // successsorParent if(successsor!=deleleNode.right){ successsorParent.left = successsor.right; successsor.right = deleleNode.right; } return successsor; } public void insert(int id){ Node newNode = new Node(id); if(root==null){ root = newNode; return; } Node current = root; Node parent = null; while(true){ parent = current; if(id