<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 Solution {
public:
    TreeNode* firstMistake = nullptr;
    TreeNode* secondMistake = nullptr;
    TreeNode* pre = nullptr;

    void recoverTree(TreeNode* root) {
        pre = new TreeNode(INT_MIN);  // Initialize pre with a minimum value node
        inorder(root);  // Perform in-order traversal to find the mistakes
        swap(firstMistake->val, secondMistake->val);  // Swap the values of the mistaken nodes
    }

    void inorder(TreeNode* root) {
        if (root == nullptr)
            return;

        inorder(root->left);  // Traverse the left subtree

        // Detect the first mistake
        if (firstMistake == nullptr && root->val < pre->val)
            firstMistake = pre;

        // Detect the second mistake
        if (firstMistake != nullptr && root->val < pre->val)
            secondMistake = root;

        pre = root;  // Update pre to the current node

        inorder(root->right);  // Traverse the right subtree
    }
};


        
   

60800