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binary-tree-inorder-traversal/README.md

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+<h2>  Binary Tree Inorder Traversal</h2><hr><div><p>Given the <code>root</code> of a binary tree, return <em>the inorder traversal of its nodes' values</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/09/15/inorder_1.jpg" style="width: 125px; height: 200px;">
+<pre><strong>Input:</strong> root = [1,null,2,3]
+<strong>Output:</strong> [1,3,2]
+</pre>
+
+<p><strong>Example 2:</strong></p>
+
+<pre><strong>Input:</strong> root = []
+<strong>Output:</strong> []
+</pre>
+
+<p><strong>Example 3:</strong></p>
+
+<pre><strong>Input:</strong> root = [1]
+<strong>Output:</strong> [1]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the tree is in the range <code>[0, 100]</code>.</li>
+	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
+</ul>
+
+<p>&nbsp;</p>
+<strong>Follow up:</strong> Recursive solution is trivial, could you do it iteratively?</div>