<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):
        i, n = 0, len(traversal)
        # define a stack
        stack = []
        num = ''

        # iterate over traversal and parse tokens
        while i < n:
            # define a depth
            level = 0

            # to calculate it for a particular (next) node
            while i < n and traversal[i] == '-': 
                i += 1
                level += 1
            
            # but if a level is different one, pop redundant
            # elements from stack
            while stack and len(stack) > level: stack.pop()

            # then we should parse an integer, since it could have
            # more than 1 digit
            while i < n and traversal[i] != '-':
                num += traversal[i]
                i += 1

            # create a Binary Tree node
            node = TreeNode(int(num))

            # and fill it with children (for left)
            if stack and stack[-1].left is None:
                stack[-1].left = node
            # for right node
            elif stack:
                stack[-1].right = node
                
            # don't forget to push a current node into stack
            stack.append(node)
            # and reset a previous integer
            num = ''
                
        # the first element of a stack is a root of a recovered tree
        return stack[0]
17300