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| if (node->left) nodes.push(node->left); | ||
| if (node->right) nodes.push(node->right); | ||
| } | ||
| result.push_back(current_level_values); |
| while (!nodes.empty()) { | ||
| int level_size = nodes.size(); | ||
| std::vector<int> current_level_values; | ||
| for (int i = 0; i < level_size; ++i) { |
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nodes の先頭 level_size 個を処理するという書き方は、 BFS の書き方としてはやや分かりにくいように感じます。 step1 のようにレベルごとに異なる配列を使うか、要素に深さも一緒に入れてあげるほうが分かりやすいと思います。
| if (!root) return {}; | ||
| std::vector<std::vector<int>> result; | ||
| std::vector<TreeNode*> nodes = {root}; // current-level nodes. | ||
| std::vector<TreeNode*> next_nodes; // next-level nodes. |
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next_nodes はループのイテレーション間で内容を持ち越さないため、スコープを短くするため、 while の中で定義したほうが良いと思いました。
| if (node->right) next_nodes.push_back(node->right); | ||
| } | ||
| result.push_back(current_level_values); | ||
| nodes = next_nodes; |
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std::swap(nodes, next_nodes) すると、コピーが発生しなくなり、やや軽くなると思います。
| class Solution { | ||
| public: | ||
| vector<vector<int>> levelOrder(TreeNode* root) { | ||
| if (!root) return {}; |
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こちらのコメントをご参照ください。
hemispherium/LeetCode_Arai60#10 (comment)
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102. Binary Tree Level Order Traversal
https://leetcode.com/problems/binary-tree-level-order-traversal/