<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]:
        tree_head = [None]
        postorder_idx = [len(postorder) - 1]

        def construct_tree_from(vals_in_subtree: List[int]) -> Optional[TreeNode]:
            '''I should be able to iterate backward in the postorder list to get
            root nodes of each subtree
            Search for the postorder value in "vals_in_subtree". Values to the left
                are in the left subtree, values to the right are in the right subtree
                Recursively call the method on these subarrays
            :return: the root of this subtree (could be None)
            :note: every call of this function increments a "global" postorder_idx'''
            if postorder_idx[0] == -1:
                # print(f'tree creation already done')
                return None
            if len(vals_in_subtree) == 0:
                return None
            
            cur_node_val = postorder[postorder_idx[0]]
            cur_node = TreeNode(cur_node_val)
            if tree_head[0] is None:
                tree_head[0] = cur_node
                # print(f'{tree_head[0]=}')
            postorder_idx[0] -= 1

            cur_val_idx = vals_in_subtree.index(cur_node_val)
            right_subtree_vals = deepcopy(vals_in_subtree[cur_val_idx + 1:])
            left_subtree_vals = deepcopy(vals_in_subtree[:cur_val_idx])

            right_subtree_root = construct_tree_from(right_subtree_vals)
            cur_node.right = right_subtree_root
            left_subtree_root = construct_tree_from(left_subtree_vals)
            cur_node.left = left_subtree_root

            return cur_node

        construct_tree_from(inorder)
        return tree_head[0]
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