Assumed Knowledge
Learning Outcomes
  • Be able to create and operate on a class holding one or more references of the same type

Author: Gaurav Gupta

The Node class

Consider the following class:

public class Node {
	public int data;
	public Node next;

	public Node(int d, Node n) {
		data = d;
		next = n;
	}
}

Every Node object holds a reference to another Node object.

Node a = new Node(10, null);

Node b = new Node(20, a);

Node head = new Node(20, new Node(10, null));

Here, we created an anonymous Node object - new Node(10, null) - and passed it as a parameter to the constructor of head.

head.next.next = new Node(-50, null);

Linking nodes

 

We can link any number of nodes as we want.

Node n5 = new Node(20, null);
Node n4 = new Node(70, n5);
Node n3 = new Node(10, n4);
Node n2 = new Node(90, n3);
Node n1 = new Node(30, n2);

Simplified representaton:

We can get all the values just using n1 :)

System.out.println(n1.data); //30
System.out.println(n1.next.data); //90
System.out.println(n1.next.next.data); //10
System.out.println(n1.next.next.next.data); //70
System.out.println(n1.next.next.next.next.data); //20

If we create a Node reference temp initialized to n1, we can re-reference it to the Node after it using temp = temp.next. Thereby, repeating the operation over and over.

Node temp = n1; //temp refers to same instance as n1
temp = temp.next; //temp refers to node after temp or n1 which is n2
temp = temp.next; //temp refers to node after temp or n2 which is n3
temp = temp.next; //temp refers to n4
temp = temp.next; //temp refers to n5
temp = temp.next; //temp is now null - STOP

Abstracting into a loop to add all the values:

Node temp = n1;
int total = 0;
while(temp != null) {
	total = total + temp.data;
	temp = temp.next;
}

Nodes can hold other objects too

In the previous example, we saw a node holding integer data, but it can hold any kind of data. For starters, take a look at RNode holding Rectangle object.

For the classes defined in,

Consider the following code,

Rectangle r1 = new Rectangle(2.5, 1.5);
Rectangle r2 = new Rectangle(4.2, 3.6);
RNode q = new RNode(r2, null);
RNode p = new RNode(r1, q);

We can create anonymous objects to reduce variable count.

RNode q = new RNode(new Rectangle(4.2, 3.6), null);
RNode p = new RNode(new Rectangle(2.5, 1.5), q);

Homework - 2

Task 1

For the class Node, draw the memory diagram to illustrate objects after the last statement of the following code executes.

Node a = new Node(20, null);
Node b = new Node(70, a);
Node c = new Node(10, a);
Node d = new Node(90, c);

Task 2

For the class Node, draw the memory diagram to illustrate objects after the last statement of the following code executes.

Node a = new Node(20, null);
Node b = new Node(70, a);
Node c = new Node(10, b);
Node d = new Node(90, c);
a.next = d;

Task 3

For the class Node, draw the memory diagram to illustrate objects after the last statement of the following code executes.

Node a = new Node(20, null);
Node b = new Node(70, a);
Node c = new Node(10, b);
Node d = new Node(90, c);
a.next = d.next;

Task 4

For the class Node, draw the memory diagram to illustrate objects after the last statement of the following code executes.

Node a = new Node(20, null);
Node b = new Node(70, a);
Node c = new Node(10, b);
Node d = new Node(90, c);
a.next = d.next.next;

Task 5

For the class Node, the following code attempts to store the sum of all items in the chain of nodes into a varaible total. However, it has a a bug. Briefly explain what is the problem with the code, and correct it.

Node a = new Node(20, null);
Node b = new Node(70, a);
Node c = new Node(10, b);
Node d = new Node(90, c);
int total = 0;
Node current = d;
while(current != null) {
	total = total + current;
	current = current.next;
}

Task 6

For the class Node, the following code attempts to store the number of nodes in the chain into a variable size. However, it has a a bug. Briefly explain what is the problem with the code, and correct it.

Node a = new Node(20, null);
Node b = new Node(70, a);
Node c = new Node(10, b);
Node d = new Node(90, c);
int size = 0;
Node current = d;
while(current != null) {
	size = size + 1;
}

Task 7

For the class Node, what is the value of result after the following code is executed?

Node a = new Node(20, null);
Node b = new Node(70, a);
Node c = new Node(10, b);
Node d = new Node(90, c);
int result = 0;
Node current = d;
while(current != null) {
	if(current.data >= 20) {
		result = result * 10 + 1;
	}
	else {
		result = result * 10;
	}
	current = current.next;
}

Task 8

For the class Node, what is the value of result after the following code executes?

Node a = new Node(9, null);
Node b = new Node(2, a);
Node c = new Node(7, b);
Node d = new Node(1, c);
a.next = d;
a.data = 	1000*d.data +
			100*d.next.data +
			10*d.next.next.data +
			1*d.next.next.next.data;

Task 9

Consider the following class definition for TreeNode:

public class TreeNode {
	public int data;
	public TreeNode left, right;
	public TreeNode(int d, TreeNode l, TreeNode r) {
		data = d;
		left = l;
		right = r;
	}
}

Draw the memory diagram to illustrate objects after the last statement of the following code executes. Also, state the number of instances and references in the diagram.

TreeNode t1 = new TreeNode(20, null, null);
TreeNode t2 = new TreeNode(-10, null, null);
TreeNode t3 = new TreeNode(70, t1, t2);