Convert binary search tree to sorted doubly linked list

We do an inorder traversal and make an arr of sorted nodes. We then create the doubly linked list by iterating through the array.

"""
# Definition for a Node.
class Node:
    def __init__(self, val, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
"""

class Solution:
    def treeToDoublyList(self, root: 'Optional[Node]') -> 'Optional[Node]':
        if not root:
            return None

        arr = []
        def inorder(root):
            if not root:
                return
            inorder(root.left)
            arr.append(root)
            inorder(root.right)
        inorder(root)

        n = len(arr)
        for i in range(n):
            arr[i].left = arr[(i-1)%n]
            arr[i].right = arr[(i+1)%n]
        arr[0].left = arr[-1]
        arr[-1].right = arr[0]

        return arr[0]

Time complexity: O(n)
Space complexity: O(n)