Exercise 2: Chained Implied Closes — Possible Solution
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THE TEST
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    parse_html("<p>A<p>B<p>C</p>")

RESULT
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    <#document>
      <p>
        'A'
      <p>
        'B'
      <p>
        'C'

Exactly THREE top-level <p> elements: "A", "B", and "C" -- each one
its own separate sibling, none nested inside another.

STACK TRACE AT EACH opentag EVENT
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Start:                          stack = [root]

1st <p> (opens):
    tok.name='p' in P_CLOSING_TAGS, but stack[-1] is root, not a p
    -> no pop. Push p1.        stack = [root, p1]
  'A' appended to p1.           p1.children = ['A']

2nd <p> (opens):
    tok.name='p' in P_CLOSING_TAGS, stack[-1]=p1, top_tag='p' -> MATCH
    -> pop p1.                  stack = [root]
    Push p2.                    stack = [root, p2]
  'B' appended to p2.           p2.children = ['B']

3rd <p> (opens):
    tok.name='p' in P_CLOSING_TAGS, stack[-1]=p2, top_tag='p' -> MATCH
    -> pop p2.                  stack = [root]
    Push p3.                    stack = [root, p3]
  'C' appended to p3.           p3.children = ['C']

</p> (closes):
    search stack for a 'p' -> stack[-1]=p3 matches -> pop.
                                 stack = [root]

Final: root.children = [p1, p2, p3]

WHY THIS CHAINS CORRECTLY WITH NO EXTRA CODE
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The check `stack[-1]` (top_tag == 'p') is re-evaluated fresh, from
scratch, every single time a new opentag token arrives -- it isn't a
one-shot check that only fires once per document. Because popping p1
and immediately pushing p2 leaves p2 sitting at the exact same "top
of stack" position p1 occupied a moment earlier, the THIRD <p>'s own
check finds p2 there and pops it using the identical logic, with no
memory of how many times this has already happened. The rule is
purely local -- "is the thing currently on top of the stack a p" --
which is exactly what makes it correctly self-chain across any
number of consecutive <p> tags without needing a counter, a loop, or
any special-casing for "this is the second time" versus "this is the
first time."