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
Intuition
Approach 1:
Brute
Sorting and comparing with actual inorder traversal
Approach 2:
3 1 2
Two cases,
Swapping are adjacent, not adjacent
Swapping are not afjacent,
Maintain the pointers
3 1 maintain 3->first, 1->middle
2 2->last
Voilation of a < b BST condition
Else if adjacent
Hence maintaing two pointers first and middle,
To swap them if at end we find last as null