Recover Binary Search Tree
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0099.Recover%20Binary%20Search%20Tree/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0099.Recover%20Binary%20Search%20Tree/images/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?
Tree; Depth-First Search; Binary Search Tree; Binary Tree
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void recoverTree(TreeNode* root) {
        TreeNode* prev = nullptr;
        TreeNode* first = nullptr;
        TreeNode* second = nullptr;
        function<void(TreeNode * root)> dfs = [&](TreeNode* root) {
            if (!root) return;
            dfs(root->left);
            if (prev && prev->val > root->val) {
                if (!first) first = prev;
                second = root;
            }
            prev = root;
            dfs(root->right);
        };
        dfs(root);
        swap(first->val, second->val);
    }
};