**BFS with level.** Every other level, reverse before appending. Or use deque: popleft vs pop for direction. O(n).
def zigzag(root):
if not root: return []
res, q, ltr = [], [root], True
while q:
level = []
for _ in range(len(q)):
n = q.pop(0)
level.append(n.val)
if n.left: q.append(n.left)
if n.right: q.append(n.right)
res += level if ltr else level[::-1]
ltr = not ltr
return res
The complete answer continues with detailed implementation patterns, architectural trade-offs, and production-grade considerations. It covers performance optimization strategies, common pitfalls to avoid, and real-world examples from companies like Meesho. The answer also includes follow-up discussion points that interviewers commonly explore.
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