Intersection Of Two Linked Lists

class ListNode:
    def __init__(self, data: int = 0, next: ListNode | None = None):
        self.data = data
        self.next = next

public class ListNode {
    int data;
    ListNode next;

    ListNode() {
        this(0, null);
    }

    ListNode(final int x) {
        this(x, null);
    }

    ListNode(final int data, final ListNode next) {
        this.data = data;
        this.next = next;
    }
}

class Solution:
    def getIntersectionNode(
        self, h1: ListNode | None, h2: ListNode | None
    ) -> ListNode | None:
        # If either of the linked list points to an empty list
        # then there can be no intersecation, do an early return
        if h1 is None or h2 is None:
            return None

        # Copy two head pointers of the linked lists into temporary variables
        p1, p2 = h1, h2

        while p1 != p2:
            p1 = p1.next
            p2 = p2.next

            # This check is needed for two equal length linked lists
            # otherwise both will reach null together and get reset to
            # the other linked list's head, creating an infinite loop
            if p1 == p2:
                break

            # Whichever pointer hits null value first, put that to the
            # head of the other linked list, essentially the idea is to
            # ensure both travel equal distant, so then converge on the
            # common first node together
            if p1 is None:
                p1 = h2

            if p2 is None:
                p2 = h1

        # It is very much possible that the two linked lists do not have
        # anything common at all, in that case, both will reach null together
        # in the second iteration, and the equality check will break the while
        # loop, so both pointers will have null, and return value will be null
        return p1

public class Solution {
    public ListNode getIntersectionNode(final ListNode h1, final ListNode h2) {
        if (null == h1 || null == h2) {
            return null;
        }

        ListNode p1 = h1;
        ListNode p2 = h2;

        while (p1 != p2) {
            p1 = p1.next;
            p2 = p2.next;

            if (p1 == p2) {
                break;
            }

            if (null == p1) {
                p1 = h2;
            }

            if (null == p2) {
                p2 = h1;
            }
        }

        return p1;
    }
}