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<h2><a href="https://leetcode.com/problems/binary-tree-postorder-traversal/">145. Binary Tree Postorder Traversal</a></h2><h3>Easy</h3><hr><div><p>Given the <code>root</code> of a&nbsp;binary tree, return <em>the postorder traversal of its nodes' values</em>.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<img alt="" src="https://assets.leetcode.com/uploads/2020/08/28/pre1.jpg" style="width: 127px; height: 200px;">
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<pre><strong>Input:</strong> root = [1,null,2,3]
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<strong>Output:</strong> [3,2,1]
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre><strong>Input:</strong> root = []
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<strong>Output:</strong> []
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre><strong>Input:</strong> root = [1]
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<strong>Output:</strong> [1]
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li>The number of the nodes in the tree is in the range <code>[0, 100]</code>.</li>
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<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
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</ul>
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<p>&nbsp;</p>
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<strong>Follow up:</strong> Recursive solution is trivial, could you do it iteratively?</div>

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