<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>

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, postorder):
        # Step 1: Create a map to store the index of each value in the inorder list
        self.inorder_map = {val: idx for idx, val in enumerate(inorder)}

        self.inorder = inorder
        self.postorder = postorder
    
        # Step 2: Initialize a helper function with the necessary parameters
        return self.buildTreeHelper(0, len(inorder) - 1)
    
    def buildTreeHelper(self, in_start, in_end):
        # Step 3: Base case when there are no elements to construct the tree
        if in_start > in_end:
            return None
        
        # Step 4: The last element in postorder is the root
        root_val = self.postorder.pop()
        root = TreeNode(root_val)
        
        # Step 5: Find the index of the root value in the inorder traversal using the hash map
        inorder_index = self.inorder_map[root_val]
        
        # Step 6: Recursively build the right and left subtrees
        # Note: Build right subtree first because postorder is being processed from the end
        root.right = self.buildTreeHelper(inorder_index + 1, in_end)
        root.left = self.buildTreeHelper(in_start, inorder_index - 1)
        
        return root

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