<p>Given two integer arrays <code>preorder</code> and <code>inorder</code> where <code>preorder</code> is the preorder traversal of a binary tree and <code>inorder</code> is the inorder traversal of the same tree, construct and return <em>the binary tree</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/tree.jpg" style="width: 277px; height: 302px;" />
<pre>
<strong>Input:</strong> preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
<strong>Output:</strong> [3,9,20,null,null,15,7]
</pre>

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

<pre>
<strong>Input:</strong> preorder = [-1], inorder = [-1]
<strong>Output:</strong> [-1]
</pre>

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

<ul>
	<li><code>1 &lt;= preorder.length &lt;= 3000</code></li>
	<li><code>inorder.length == preorder.length</code></li>
	<li><code>-3000 &lt;= preorder[i], inorder[i] &lt;= 3000</code></li>
	<li><code>preorder</code> and <code>inorder</code> consist of <strong>unique</strong> values.</li>
	<li>Each value of <code>inorder</code> also appears in <code>preorder</code>.</li>
	<li><code>preorder</code> is <strong>guaranteed</strong> to be the preorder traversal of the tree.</li>
	<li><code>inorder</code> is <strong>guaranteed</strong> to be the inorder traversal of the tree.</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 buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        if not preorder or not inorder:
            return None
        
        root = TreeNode(preorder[0])

        def pre(curr, left, right, pre_index):
            if pre_index >= len(preorder):
                return pre_index  # Safeguard against index out of bounds
            
            # If left and right subtrees are empty, return pre_index
            if not left and not right:
                return pre_index

            pre_val = preorder[pre_index]
            
            # Process left subtree if pre_val is in the left subtree
            if pre_val in left:
                new_l = TreeNode(pre_val)
                curr.left = new_l
                l_ind = left.index(pre_val)
                left_subtree = safe_slice(left, 0, l_ind)
                right_subtree = safe_slice(left, l_ind + 1, len(left))

                pre_index = pre(curr.left, left_subtree, right_subtree, pre_index + 1)
            
            if pre_index < len(preorder):  # Ensure pre_index is still valid
                pre_val = preorder[pre_index]

                # Process right subtree if pre_val is in the right subtree
                if pre_val in right:
                    new_r = TreeNode(pre_val)
                    curr.right = new_r
                    r_ind = right.index(pre_val)
                    left_subtree = safe_slice(right, 0, r_ind)
                    right_subtree = safe_slice(right, r_ind + 1, len(right))

                    pre_index = pre(curr.right, left_subtree, right_subtree, pre_index + 1)
            
            return pre_index  # Return the updated pre_index

        # Helper function to safely slice the list
        def safe_slice(lst, start, end):
            if start >= len(lst) or start < 0 or end < 0 or start >= end:
                return []
            return lst[start:end]

        # Split the inorder list into left and right subtrees
        index = inorder.index(preorder[0])
        left = inorder[:index]
        right = inorder[index + 1:]

        # Start recursive construction of the tree
        pre(root, left, right, 1)

        return root
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