<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 __init__(self):
        self.root = None
        self.count = 0

    def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        return self.build(preorder, inorder, None)

    def build(self, preorder, inorder, curr):
        left_preorder = []
        left_inorder = []

        right_preorder = []
        right_inorder = []
        
        if len(preorder) == 0:
            return

        cur_node = preorder.pop(0)
        left_nodes = inorder.pop(0)
        while left_nodes != cur_node:
            left_preorder.append(preorder.pop(0))
            left_inorder.append(left_nodes)
            left_nodes = inorder.pop(0)


        while len(preorder) > 0:
            right_preorder.append(preorder.pop(0))
            right_inorder.append(inorder.pop(0))

        
        if self.count == 0:
            curr = TreeNode(cur_node)
            self.root = curr
            
        self.count += 1

        left = None
        if not len(left_preorder) == 0:
            left = TreeNode(left_preorder[0])
            curr.left = left

        right = None
        if not len(right_preorder) == 0:
            right = TreeNode(right_preorder[0])
            curr.right = right
        

        
        self.build(right_preorder, right_inorder, right)
        self.build(left_preorder, left_inorder, left)


        return self.root

            

        
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