<p>Given the <code>root</code> node of a binary tree, your task is to create a string representation of the tree following a specific set of formatting rules. The representation should be based on a preorder traversal of the binary tree and must adhere to the following guidelines:</p>

<ul>
	<li>
	<p><strong>Node Representation</strong>: Each node in the tree should be represented by its integer value.</p>
	</li>
	<li>
	<p><strong>Parentheses for Children</strong>: If a node has at least one child (either left or right), its children should be represented inside parentheses. Specifically:</p>

	<ul>
		<li>If a node has a left child, the value of the left child should be enclosed in parentheses immediately following the node&#39;s value.</li>
		<li>If a node has a right child, the value of the right child should also be enclosed in parentheses. The parentheses for the right child should follow those of the left child.</li>
	</ul>
	</li>
	<li>
	<p><strong>Omitting Empty Parentheses</strong>: Any empty parentheses pairs (i.e., <code>()</code>) should be omitted from the final string representation of the tree, with one specific exception: when a node has a right child but no left child. In such cases, you must include an empty pair of parentheses to indicate the absence of the left child. This ensures that the one-to-one mapping between the string representation and the original binary tree structure is maintained.</p>

	<p>In summary, empty parentheses pairs should be omitted when a node has only a left child or no children. However, when a node has a right child but no left child, an empty pair of parentheses must precede the representation of the right child to reflect the tree&#39;s structure accurately.</p>
	</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/03/cons1-tree.jpg" style="padding: 10px; background: #fff; border-radius: .5rem;" />
<pre>
<strong>Input:</strong> root = [1,2,3,4]
<strong>Output:</strong> &quot;1(2(4))(3)&quot;
<strong>Explanation:</strong> Originally, it needs to be &quot;1(2(4)())(3()())&quot;, but you need to omit all the empty parenthesis pairs. And it will be &quot;1(2(4))(3)&quot;.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/05/03/cons2-tree.jpg" style="padding: 10px; background: #fff; border-radius: .5rem;" />
<pre>
<strong>Input:</strong> root = [1,2,3,null,4]
<strong>Output:</strong> &quot;1(2()(4))(3)&quot;
<strong>Explanation:</strong> Almost the same as the first example, except the <code>()</code> after <code>2</code> is necessary to indicate the absence of a left child for <code>2</code> and the presence of a right child.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 10<sup>4</sup>]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
</ul>

# 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 tree2str(self, root: Optional[TreeNode]) -> str:
        res = []
        def preorder(root):
            if not root:
                return
            res.append("(")
            res.append(str(root.val))
            if not root.left and root.right:
                res.append("()")
            preorder(root.left)
            preorder(root.right)
            res.append(")")
        preorder(root)
        return "".join(res)[1:-1]
18100