Recover Binary Search Tree

You are given the root of a binary search tree (BST), where the values of exactly two nodes of the tree were swapped by mistake. Recover the tree without changing its structure.

Example 1:

Input: root = [1,3,null,null,2]
Output: [3,1,null,null,2]
Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid.

Example 2:

Input: root = [3,1,4,null,null,2]
Output: [2,1,4,null,null,3]
Explanation: 2 cannot be in the right subtree of 3 because 2 < 3. Swapping 2 and 3 makes the BST valid.

Constraints:

• The number of nodes in the tree is in the range [2, 1000].
• -231 <= Node.val <= 231 - 1

Follow up: A solution using O(n) space is pretty straight-forward. Could you devise a constant O(1) space solution?SOLUTION:

# 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 inorder(self, root, append):
        if root:
            self.inorder(root.left, append)
            if append:
                self.vals.append(root.val)
            else:
                root.val = self.vals[self.i]
                self.i += 1
            self.inorder(root.right, append)

    def recoverTree(self, root: Optional[TreeNode]) -> None:
        self.vals = []
        self.inorder(root, True)
        self.vals.sort()
        self.i = 0
        self.inorder(root, False)