Construct Binary Tree from Preorder and Inorder Traversal
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0105.Construct%20Binary%20Tree%20from%20Preorder%20and%20Inorder%20Traversal/images/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>
Tree; Array; Hash Table; Divide and Conquer; Binary Tree
# 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 dfs(i: int, j: int, n: int) -> Optional[TreeNode]:
            if n <= 0:
                return None
            v = preorder[i]
            k = d[v]
            l = dfs(i + 1, j, k - j)
            r = dfs(i + 1 + k - j, k + 1, n - k + j - 1)
            return TreeNode(v, l, r)

        d = {v: i for i, v in enumerate(inorder)}
        return dfs(0, 0, len(preorder))
