<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]:
        # in order: left node right
        # left right node
        in_dict = {elem:i for i, elem in enumerate(inorder)}
        post_dict = {elem:i for i, elem in enumerate(postorder)}
        print(post_dict)
        # now knowing this: we know the first most node is the left most in in order and post order
        # then it's the root:

        # so strategy should be start at node find corresponding index in in order
        # 3, 9, 1
        root = TreeNode(postorder[-1], None, None)
        # right s
        # because everything right
        # [12, 10, 9, 11, 3, 15, 20, 7]
        # [12, 10, 11, 9, 15, 7, 20, 3]

        # we use post to answer what's the root?
        # latest node is
        def construct_tree(post_ind, left, right):
            curr_elem = postorder[post_ind]

            curr_node = TreeNode(curr_elem, None, None)
            if left == right:
                return curr_node
            
            in_ind = in_dict[curr_elem]

            # next_post_index

            # everything between left and in_ind is the left tree
            # everything between in_ind + 1 and right is the right tree
            left_nodes = inorder[left:in_ind]
            right_nodes = inorder[in_ind + 1:right + 1]

            left_post_ind = -1

            for left_node in left_nodes:
                left_post_ind = max(left_post_ind, post_dict[left_node])
            
            right_post_ind = -1

            for right_node in right_nodes:
                right_post_ind = max(right_post_ind, post_dict[right_node])
            
            if left_post_ind != -1:
                curr_node.left = construct_tree(left_post_ind, left, in_ind - 1)
            
            if right_post_ind != -1:
                curr_node.right = construct_tree(right_post_ind, in_ind + 1, right)

            return curr_node
        
        tree = construct_tree(len(postorder) - 1, 0, len(inorder) - 1)
        return tree






            


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