<p>Given two integer arrays <code>preorder</code> and <code>inorder</code> where <code>preorder</code> is the preorder traversal of a binary tree and <code>inorder</code> is the inorder 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> preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
<strong>Output:</strong> [3,9,20,null,null,15,7]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> preorder = [-1], inorder = [-1]
<strong>Output:</strong> [-1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= preorder.length &lt;= 3000</code></li>
	<li><code>inorder.length == preorder.length</code></li>
	<li><code>-3000 &lt;= preorder[i], inorder[i] &lt;= 3000</code></li>
	<li><code>preorder</code> and <code>inorder</code> consist of <strong>unique</strong> values.</li>
	<li>Each value of <code>inorder</code> also appears in <code>preorder</code>.</li>
	<li><code>preorder</code> is <strong>guaranteed</strong> to be the preorder traversal of the tree.</li>
	<li><code>inorder</code> is <strong>guaranteed</strong> to be the inorder 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, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
        index_look_up = {inorder[i]:i for i in range(len(inorder))}
        root_val = preorder.pop(0)
        root = TreeNode(root_val)
        left_sub = (0, index_look_up[root_val]) 
        right_sub = (index_look_up[root_val]+1, len(inorder))

        def build(parent, left_sub, right_sub):
            left_sub_tree = inorder[left_sub[0]: left_sub[1]]
            right_sub_tree = inorder[right_sub[0]: right_sub[1]]
            if len(preorder) == 0 or (len(left_sub_tree)==0 and len(right_sub_tree)==0):
                return
            
            next_val = preorder.pop(0)
            if next_val in left_sub_tree:
                # left sub tree build
                new_node = TreeNode(next_val)
                parent.left = new_node
                build(new_node, (left_sub[0], index_look_up[next_val]), (index_look_up[next_val]+1,left_sub[1]))
            
                # right sub tree build
                if len(preorder) == 0 or len(right_sub_tree)==0:
                    return
                
                next_val = preorder.pop(0)
                assert next_val in right_sub_tree
                new_node = TreeNode(next_val)
                parent.right = new_node
                build(new_node, (right_sub[0], index_look_up[next_val]), (index_look_up[next_val]+1,right_sub[1]))
            
            elif next_val in right_sub_tree:
                # right sub tree build
                new_node = TreeNode(next_val)
                parent.right = new_node
                build(new_node, (right_sub[0], index_look_up[next_val]), (index_look_up[next_val]+1,right_sub[1]))

            return
        
        build(root, left_sub, right_sub)

        return root
                
                

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