338. Familystrokes Review

internalCnt ← 0 // |I| horizontalCnt ← 0 // # childCount(v) ≥ 2

int main() I import sys sys.setrecursionlimit(200000) 338. FamilyStrokes

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 internalCnt ← 0 // |I| horizontalCnt ← 0

print(internal + horizontal)

if childCnt > 0: // v has at least one child → internal internalCnt += 1 if childCnt >= 2: horizontalCnt += 1 6.1 C++17 #include &lt

Both bounds comfortably meet the limits for N ≤ 10⁵ . Below are clean, self‑contained implementations in C++17 and Python 3 that follow the algorithm exactly. 6.1 C++17 #include <bits/stdc++.h> using namespace std;

Proof. If childCnt ≥ 2 : the children occupy at least two columns on the next row, so a horizontal line is needed to connect the leftmost to the rightmost child (rule 2).