No. It may need to try other children. It can stop as soon as one succeeds because this query asks for existence, but one failed choice does not prove the other choices fail.

def matches(node, pattern, i=0):
    if i == len(pattern):
        return None in node
    char = pattern[i]
    if char == ".":
        return any(matches(child, pattern, i + 1)
                   for key, child in node.items() if key is not None)
    return char in node and matches(node[char], pattern, i + 1)

trie = {"c": {"a": {"r": {None: True}, "t": {None: True}}}}
assert matches(trie, "ca.")
assert not matches(trie, "c.")
assert not matches(trie, "dog")

The state is the current trie node and position in the pattern. No mutation is needed, so recursive return already isolates alternatives. An ordinary character follows at most one edge. A wildcard follows all character edges until a successful completion is found.

For a trie with N nodes and pattern length L, this simple fixed-length wildcard search visits at most O(N) nodes, restricted to depth L, and uses O(L) recursion space. It can inspect far more nodes than a literal lookup.

Transfer to a letter board: explore adjacent cells while following trie edges. A missing prefix edge prunes the entire branch. Now a cell cannot usually repeat within the current word, so board search also needs path-local visited information that is restored when exploring another route. A global visited-cell set would lose words.

Analogy limit: ., matching exactly one character, differs from *, which may match several. Changing wildcard semantics changes transitions, state dependencies, and possibly repeated states.

No. Explore all matching branches and construct their outputs. The data structure can prune impossible prefixes, but it cannot skip required matches.