<p>We run a&nbsp;preorder&nbsp;depth-first search (DFS) on the <code>root</code> of a binary tree.</p>

<p>At each node in this traversal, we output <code>D</code> dashes (where <code>D</code> is the depth of this node), then we output the value of this node.&nbsp; If the depth of a node is <code>D</code>, the depth of its immediate child is <code>D + 1</code>.&nbsp; The depth of the <code>root</code> node is <code>0</code>.</p>

<p>If a node has only one child, that child is guaranteed to be <strong>the left child</strong>.</p>

<p>Given the output <code>traversal</code> of this traversal, recover the tree and return <em>its</em> <code>root</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/09/10/recover_tree_ex1.png" style="width: 423px; height: 200px;" />
<pre>
<strong>Input:</strong> traversal = &quot;1-2--3--4-5--6--7&quot;
<strong>Output:</strong> [1,2,5,3,4,6,7]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/09/10/recover_tree_ex2.png" style="width: 432px; height: 250px;" />
<pre>
<strong>Input:</strong> traversal = &quot;1-2--3---4-5--6---7&quot;
<strong>Output:</strong> [1,2,5,3,null,6,null,4,null,7]
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2024/09/10/recover_tree_ex3.png" style="width: 305px; height: 250px;" />
<pre>
<strong>Input:</strong> traversal = &quot;1-401--349---90--88&quot;
<strong>Output:</strong> [1,401,null,349,88,90]
</pre>

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

<ul>
	<li>The number of nodes in the original tree is in the range <code>[1, 1000]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def recoverFromPreorder(self, traversal: str) -> Optional[TreeNode]:

        def build(level): 
            nonlocal index 
            if level > values[index]: 
                return
            node = TreeNode(int(values[index+1]))
            index += 2 
            node.left = build(level+1)
            node.right = build(level+1)
            return node 
        
        index = 0 
        values = [0]+[(val, len(val))[i&1] for i,val in enumerate(re.findall('-+|\d+', traversal))]+[-1]
        
        return build(0)
17600