Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 16.9 MB (94.54%)

Author: koralkulacoglu

LeetCode/0102-binary-tree-level-order-traversal/README.md

@@ -0,0 +1,31 @@
+<p>Given the <code>root</code> of a binary tree, return <em>the level order traversal of its nodes&#39; values</em>. (i.e., from left to right, level by level).</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg" style="width: 277px; height: 302px;" />
+<pre>
+<strong>Input:</strong> root = [3,9,20,null,null,15,7]
+<strong>Output:</strong> [[3],[9,20],[15,7]]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> root = [1]
+<strong>Output:</strong> [[1]]
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> root = []
+<strong>Output:</strong> []
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the tree is in the range <code>[0, 2000]</code>.</li>
+	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
+</ul>

LeetCode/0102-binary-tree-level-order-traversal/solution.cpp

@@ -0,0 +1,29 @@
+/**
+ * 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>> res;
+        if (root == nullptr) return res;
+        queue<pair<TreeNode*, int>> q;
+        q.push({root, 0});
+        while (!q.empty()) {
+            auto [node, level] = q.front();
+            q.pop();
+            if (level == res.size()) res.push_back({});
+            res[level].push_back(node->val);
+            if (node->left != nullptr) q.push({node->left, level+1});
+            if (node->right != nullptr) q.push({node->right, level+1});
+        }
+        return res;
+    }
+};