<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://assets.leetcode.com/uploads/2021/02/19/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>

# 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]:
        if len(inorder) != len(postorder):
            return None

        in_map = {val: idx for idx, val in enumerate(inorder)}
        return self._buildTree(inorder, 0, len(inorder) - 1, postorder, 0, len(postorder) - 1, in_map)

    def _buildTree(self, inorder, in_start, in_end, postorder, post_start, post_end, in_map):
        if in_start > in_end or post_start > post_end:
            return None

        # The last element in postorder is the root of the tree/subtree
        root_val = postorder[post_end]
        root = TreeNode(root_val)

        # Find the index of the root in the inorder list
        in_root_idx = in_map[root_val]

        # Number of nodes in the left subtree
        nums_left = in_root_idx - in_start

        # Recursively build the left and right subtrees
        root.left = self._buildTree(inorder, in_start, in_root_idx - 1,
                                    postorder, post_start, post_start + nums_left - 1, in_map)
        root.right = self._buildTree(inorder, in_root_idx + 1, in_end,
                                     postorder, post_start + nums_left, post_end - 1, in_map)
        return root

        in_map = {val: idx for idx, val in enumerate(inorder)}
        return self._buildTree(inorder, 0, len(inorder) - 1, postorder, 0, len(postorder) - 1, in_map)

    def _buildTree(self, inorder: List[int], in_start: int, in_end: int, postorder: List[int], post_start: int, post_end: int, in_map: dict) -> Optional[TreeNode]:
        if in_start > in_end or post_start > post_end:
            return None

        # The last element in postorder is the root of the tree/subtree
        root_val = postorder[post_end]
        root = TreeNode(root_val)

        # Find the index of the root in the inorder list
        in_root_idx = in_map[root_val]

        # Number of nodes in the left subtree
        nums_left = in_root_idx - in_start

        # Recursively build the left and right subtrees
        root.left = self._buildTree(inorder, in_start, in_root_idx - 1,
                                    postorder, post_start, post_start + nums_left - 1, in_map)
        root.right = self._buildTree(inorder, in_root_idx + 1, in_end,
                                     postorder, post_start + nums_left, post_end - 1, in_map)
        return root
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