102. Binary Tree Level Order Traversal

0102-binary-tree-level-order-traversal/memo.md

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+### step1
+
+dfsで各ノードを走査しながら、もっている各階層のlevelのvector[level]にnodeをpush_backしていく。
+
+### step2
+
+名前がてきとうだったのでましな名前に変更。
+
+### step3
+
+ChatGPTに聞いたところ、「👉 レベル順探索 = BFSが本質的に一致」ということがわかったのでBFSで実装し直し。
+3回通すまで書き直し。

0102-binary-tree-level-order-traversal/step1.cpp

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+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    void dfs(TreeNode* root, int level, vector<vector<int>>& vvi) {
+        if (!root) {
+            return;
+        }
+        if (vvi.size() < (level + 1)) {
+            vvi.resize(level + 1);
+        }
+        vvi[level].push_back(root->val);
+        if (root->left) {
+            dfs(root->left, level + 1, vvi);
+        }
+        if (root->right) {
+            dfs(root->right, level + 1, vvi);
+        }
+    }
+
+    vector<vector<int>> levelOrder(TreeNode* root) {
+        int level = 0;
+        vector<vector<int>> vvi;
+        dfs(root, level, vvi);
+        return vvi;
+    }
+};

0102-binary-tree-level-order-traversal/step2.cpp

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+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    void scan(TreeNode* root, int level, vector<vector<int>>& nodes) {
+        if (!root) {
+            return;
+        }
+        if (nodes.size() < (level + 1)) {
+            nodes.resize(level + 1);
+        }
+        nodes[level].push_back(root->val);
+        if (root->left) {
+            scan(root->left, level + 1, nodes);
+        }
+        if (root->right) {
+            scan(root->right, level + 1, nodes);
+        }
+    }
+
+    vector<vector<int>> levelOrder(TreeNode* root) {
+        int level = 0;
+        vector<vector<int>> nodes;
+        scan(root, level, nodes);
+        return nodes;
+    }
+};

0102-binary-tree-level-order-traversal/step3.cpp

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+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    vector<vector<int>> levelOrder(TreeNode* root) {
+        vector<vector<int>> result;
+        if (!root) return result;
+
+        queue<TreeNode*> nodes;
+        nodes.push(root);
+
+        while (!nodes.empty()) {
+            int size = nodes.size();
+            vector<int> level;
+
+            for (int i = 0; i < size; i++) {
+                TreeNode* node = nodes.front();
+                nodes.pop();
+
+                level.push_back(node->val);
+
+                if (node->left) nodes.push(node->left);
+                if (node->right) nodes.push(node->right);
+            }
+
+            result.push_back(level);
+        }
+
+        return result;
+    }
+};