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mamo3gr
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Mar 24, 2026
| next_frontier = deque([]) | ||
| values_at_this_level = [] | ||
| while frontier: | ||
| node = frontier.popleft() |
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for node in frontier でループを回せば、deque が不要になりそうです。
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https://leetcode.com/problems/binary-tree-level-order-traversal/description/