如何序列化在java中实现的链表？

public class ListNode implements Serializable {
    private int val;
    private ListNode next;

    public ListNode(int val) {
        this.val = val;
    }

    public int getVal() {
        return val;
    }

    public void setVal(int val) {
        this.val = val;
    }

    public ListNode getNext() {
        return next;
    }

    public void setNext(ListNode next) {
        this.next = next;
    }
}

import java.io.*;

public class SerializeLinkedList {
    public static void main(String[] args) {
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(3);

        try {
            FileOutputStream fileOut = new FileOutputStream("linkedList.ser");
            ObjectOutputStream out = new ObjectOutputStream(fileOut);
            out.writeObject(head);
            out.close();
            fileOut.close();
        } catch (IOException i) {
            i.printStackTrace();
        }
    }
}

import java.io.*;

public class DeserializeLinkedList {
    public static void main(String[] args) {
        ListNode head = null;
        try {
            FileInputStream fileIn = new FileInputStream("linkedList.ser");
            ObjectInputStream in = new ObjectInputStream(fileIn);
            head = (ListNode) in.readObject();
            in.close();
            fileIn.close();
        } catch (IOException i) {
            i.printStackTrace();
        } catch (ClassNotFoundException c) {
            System.out.println("ListNode class not found");
            c.printStackTrace();
        }

        ListNode currentNode = head;
        while (currentNode != null) {
            System.out.println(currentNode.getVal());
            currentNode = currentNode.getNext();
        }
    }
}