<p>Given the <code>root</code> of a binary tree, return <em>the preorder traversal of its nodes&#39; values</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">root = [1,null,2,3]</span></p>

<p><strong>Output:</strong> <span class="example-io">[1,2,3]</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/29/screenshot-2024-08-29-202743.png" style="width: 200px; height: 264px;" /></p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">root = [1,2,3,4,5,null,8,null,null,6,7,9]</span></p>

<p><strong>Output:</strong> <span class="example-io">[1,2,4,5,6,7,3,8,9]</span></p>

<p><strong>Explanation:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/29/tree_2.png" style="width: 350px; height: 286px;" /></p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">root = []</span></p>

<p><strong>Output:</strong> <span class="example-io">[]</span></p>
</div>

<p><strong class="example">Example 4:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">root = [1]</span></p>

<p><strong>Output:</strong> <span class="example-io">[1]</span></p>
</div>

<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>
<p><strong>Follow up:</strong> Recursive solution is trivial, could you do it iteratively?</p>

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        result = []
        def helper(node):
            if not node:
                return node
            result.append(node.val)

            helper(node.left)
            helper(node.right)

        helper(root)
        return result

        


        
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