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Abhishek Chaudhary

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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)
``````