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| if root == nil { | ||
| return [][]int{} | ||
| } | ||
| nodeValuesByLevel := make([][]int, 0, 2000) |
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ちょうど先ほど別問題で似たような指摘を受けました
定数にする or 定めない の選択肢がありますが、最大要素数が2000程度ならそんなに実行時間の短縮に効果が期待できないのでキャパシティを定めなくていいかなと思いました
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https://leetcode.com/problems/binary-tree-level-order-traversal/description/