Construct Binary Tree from Inorder and Postorder Traversal
<p>Given two integer arrays <code>inorder</code> and <code>postorder</code> where <code>inorder</code> is the inorder traversal of a binary tree and <code>postorder</code> is the postorder 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/0100-0199/0106.Construct%20Binary%20Tree%20from%20Inorder%20and%20Postorder%20Traversal/images/tree.jpg" style="width: 277px; height: 302px;" />
<pre>
<strong>Input:</strong> inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
<strong>Output:</strong> [3,9,20,null,null,15,7]
</pre>

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

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

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

<ul>
	<li><code>1 &lt;= inorder.length &lt;= 3000</code></li>
	<li><code>postorder.length == inorder.length</code></li>
	<li><code>-3000 &lt;= inorder[i], postorder[i] &lt;= 3000</code></li>
	<li><code>inorder</code> and <code>postorder</code> consist of <strong>unique</strong> values.</li>
	<li>Each value of <code>postorder</code> also appears in <code>inorder</code>.</li>
	<li><code>inorder</code> is <strong>guaranteed</strong> to be the inorder traversal of the tree.</li>
	<li><code>postorder</code> is <strong>guaranteed</strong> to be the postorder 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, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
        def dfs(i: int, j: int, n: int) -> Optional[TreeNode]:
            if n <= 0:
                return None
            v = postorder[j + n - 1]
            k = d[v]
            l = dfs(i, j, k - i)
            r = dfs(k + 1, j + k - i, n - k + i - 1)
            return TreeNode(v, l, r)

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