In preorder traversal, the first element is always the root. We find this element's position in the inorder traversal. All elements to the left in inorder are in the left subtree, and all elements to the right are in the right subtree.

1. Base case: If preorder or inorder is empty, return None.
2. root_val = preorder[0].
3. Create root = TreeNode(root_val).
4. Find index of root_val in inorder using a map for O(1) lookups.
5. Recursively build left and right subtrees:
   • Left subtree uses inorder[:index] and preorder[1:index+1].
   • Right subtree uses inorder[index+1:] and preorder[index+1:].

• Time Complexity: O(N), using a hash map to find indices.
• Space Complexity: O(N) to store the hash map and nodes.

def build_tree(preorder, inorder):
    in_map = {val: i for i, val in enumerate(inorder)}
    pre_idx = 0
    
    def helper(left, right):
        nonlocal pre_idx
        if left > right: return None
        
        root_val = preorder[pre_idx]
        root = TreeNode(root_val)
        pre_idx += 1
        
        # Root splits inorder into left and right subtrees
        mid = in_map[root_val]
        root.left = helper(left, mid - 1)
        root.right = helper(mid + 1, right)
        
        return root
        
    return helper(0, len(inorder) - 1)