<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 not inorder:
            return

        if len(inorder) == 1:
            return TreeNode(inorder[0])

        root_val = postorder[-1]
        root = TreeNode(root_val)

        rootindex = inorder.index(root_val)

        root.left = self.buildTree(inorder[:rootindex], postorder[:rootindex])
        root.right = self.buildTree(inorder[rootindex+1:], postorder[rootindex:-1])

        return root
        '''
        '''
        遞迴調用的深度：
遞迴的深度在最壞情況下是 n（當樹退化成鏈表時），每次遞迴都會佔用記憶體堆疊。這部分空間複雜度是 O(n)。
切片創建的新列表：
每次遞迴調用會創建新的子列表，這些切片操作佔用了額外的空間。最壞情況下，每次創建的子列表的大小是 O(n)，而且每次遞迴都需要這樣的額外空間。因此，切片操作的空間複雜度是 O(n^2) 
        time also O(n^2) 每次調用 index 都需要遍歷 inorder 列表，這是一個 O(n) 的操作，
        '''

        inorder_map = {val: index for index, val in enumerate(inorder)}

        def helper(il: int, ir: int, pl: int, pr: int):
            if ir < il or pr < pl:
                return None
            
            rootval = postorder[pr]
            root = TreeNode(rootval)

            rootindex = inorder_map[rootval]

            lefttreesize = rootindex - il

            root.left = helper(il, rootindex-1, pl, pl+lefttreesize-1)
            root.right = helper(rootindex+1, ir, pl+lefttreesize, pr-1)

            return root

        return helper(0, len(inorder)-1, 0, len(postorder)-1)


        
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