<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]:

        def recurse(preorder, inorder, preorder_ind: list[int]):

            # base cases
            if len(inorder) == 1: # leaf node
                root_obj = TreeNode(preorder[preorder_ind[0]])
                preorder_ind[0] += 1
                return root_obj
            elif len(inorder) == 0: # no node
                return None

            # first node of "preorder" is the root
            root = preorder[preorder_ind[0]]
            root_obj = TreeNode(root)
            preorder_ind[0] += 1


            # 1. find the root in inorder
            for i, node in enumerate(inorder):
                if node == root:
                    # 2. recurse on left subtree
                    root_obj.left = recurse(preorder, inorder[:i], preorder_ind)
                    break
            

            # 3. recurse on right subtree
            root_obj.right = recurse(preorder, inorder[i+1:], preorder_ind)

            return root_obj
        
        return recurse(preorder, inorder, [0])

39890