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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/28/recover1.jpg" style="width: 422px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [1,3,null,null,2]
<strong>Output:</strong> [3,1,null,null,2]
<strong>Explanation:</strong> 3 cannot be a left child of 1 because 3 &gt; 1. Swapping 1 and 3 makes the BST valid.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/28/recover2.jpg" style="width: 581px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [3,1,4,null,null,2]
<strong>Output:</strong> [2,1,4,null,null,3]
<strong>Explanation:</strong> 2 cannot be in the right subtree of 3 because 2 &lt; 3. Swapping 2 and 3 makes the BST valid.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[2, 1000]</code>.</li>
	<li><code>-2<sup>31</sup> &lt;= Node.val &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> A solution using <code>O(n)</code> space is pretty straight-forward. Could you devise a constant <code>O(1)</code> space solution?
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def __init__(self):
        self.first = None
        self.second = None
        self.prev = None

    def recoverTree(self, root: TreeNode) -> None:
        # Helper function for inorder traversal
        def inorder(node: TreeNode):
            if not node:
                return
            
            # Traverse the left subtree
            inorder(node.left)
            
            # Check if we found a swapped node
            if self.prev and self.prev.val > node.val:
                # If it's the first occurrence, set the first node
                if not self.first:
                    self.first = self.prev
                # Always set the second node
                self.second = node
            
            # Update prev pointer
            self.prev = node
            
            # Traverse the right subtree
            inorder(node.right)
        
        # Start inorder traversal to identify swapped nodes
        inorder(root)
        
        # Swap the values of the two nodes
        if self.first and self.second:
            self.first.val, self.second.val = self.second.val, self.first.val

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