99-recover-binary-search-tree.py

"""
Problem Link: https://leetcode.com/problems/recover-binary-search-tree/

You are given the root of a binary search tree (BST), where exactly two nodes of the tree were swapped by mistake. 
Recover the tree without changing its structure.
Follow up: A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

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
"""# 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 recoverTree(self, root: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        stack = []
        cur = root
        prev = None
        first = second = None
        
        while stack or cur:
            if cur:
                stack.append(cur)
                cur = cur.left
            else:
                node = stack.pop()
                
                if prev and prev.val >= node.val:
                    if not first:
                        first = prev
                    
                    if first:
                        second  = node
                    
                prev = node
                cur = node.right
                
        first.val, second.val = second.val, first.val