if childCnt > 0: // v has at least one child → internal internalCnt += 1 if childCnt >= 2: horizontalCnt += 1
while stack: v, p = stack.pop() child_cnt = 0 for w in g[v]: if w == p: continue child_cnt += 1 stack.append((w, v)) if child_cnt: internal += 1 if child_cnt >= 2: horizontal += 1 338. FamilyStrokes
def main() -> None: data = sys.stdin.read().strip().split() if not data: return it = iter(data) n = int(next(it)) g = [[] for _ in range(n + 1)] for _ in range(n - 1): u = int(next(it)); v = int(next(it)) g[u].append(v) g[v].append(u) if childCnt > 0: // v has at
root = 1 stack = [(root, 0)] # (node, parent) internal = 0 horizontal = 0 ∎ An internal node requires a horizontal stroke
Proof. The drawing rules require a vertical line from the node down to the row of its children whenever it has at least one child. The line is mandatory and unique, hence exactly one vertical stroke. ∎ An internal node requires a horizontal stroke iff childCnt ≥ 2 .
Memory – The adjacency list stores 2·(N‑1) integers, plus a stack/queue of at most N entries and a few counters: O(N) .
Only‑if childCnt = 1 : the sole child is placed directly under the parent; the horizontal segment would have length zero and is omitted by the drawing convention. ∎ The number of strokes contributed by a node v is