📦 Adib234 / Leetcode

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24# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
        if headA == None or headB == None:
            return None
        a = headA
        b = headB

        while a != b:
            if a == None:
                a = headB
            else:
                a = a.next
            if b == None:
                b = headA
            else:
                b = b.next
        return a