No. The minimum remaining node must be one of two current fronts.

Each chain offers its smallest remaining value at the front. The smaller front is globally safe; attach it and repeat the same question.

What the function promises: Return a nondecreasing chain reusing nodes. Inputs are disjoint, acyclic, and sorted.

class Node:
    def __init__(self, value, next_node=None):
        self.value, self.next = value, next_node

def build(values):
    head = None
    for value in reversed(values):
        head = Node(value, head)
    return head

def to_list(head):
    answer = []
    while head is not None:
        answer.append(head.value)
        head = head.next
    return answer

def merge_lists(left, right):
    dummy = tail = Node(None)
    while left is not None and right is not None:
        if left.value <= right.value:
            tail.next, left = left, left.next
        else:
            tail.next, right = right, right.next
        tail = tail.next
    tail.next = left if left is not None else right
    return dummy.next

assert to_list(merge_lists(build([1,4]), build([2,3]))) == [1,2,3,4]

Work and storage: O(n+m) time and O(1) auxiliary space.

When to recognize it: The minimum remaining node must be one of two current fronts.

Check before applying it: Reused nodes mutate the input chains; shared nodes or cycles violate the simple ownership contract.

Keep the current front of each nonempty list in a min-heap.