338. Familystrokes <TOP-RATED>

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).

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; 338. FamilyStrokes

while stack not empty: v, p = pop(stack) childCnt = 0 for each w in G[v]: if w == p: continue // ignore the edge back to parent childCnt += 1 push (w, v) on stack Below are clean

int main() long long horizontalCnt = 0; // # childCnt >= 2 using namespace std

print(internal + horizontal)

root = 1 stack = [(root, 0)] # (node, parent) internal = 0 horizontal = 0