<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]:
        self.in_map = {elem: idx for idx, elem in enumerate(inorder)}

        first = preorder[0]

        return self.rec_get_tree(first, -1, float("inf"), preorder[1:])

    def rec_get_tree(self, val: int, lower, upper, preorder: list[int]):
        val_idx = self.in_map[val]

        left_node = None
        right_node = None

        for idx, elem in enumerate(preorder):
            elem_map_val = self.in_map[elem]

            if not lower < elem_map_val < upper:
                continue

            lt = elem_map_val < val_idx
            
            if (not left_node) and lt:
                func = lambda x: (x < val_idx) and parent_check(x)
                left_node = self.rec_get_tree(elem, lower, val_idx, preorder[idx+1:])
                
            elif (not right_node) and (not lt):
                func = lambda x: (x > val_idx) and parent_check(x)
                right_node = self.rec_get_tree(elem, val_idx, upper, preorder[idx+1:])

            if left_node and right_node:
                break

        return TreeNode(
            val=val, 
            left=left_node,
            right=right_node
        )

        
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