题目：
Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

解答：
想法很简单，找到两个顺序不对的点，交换。但是如何知道这两个点的顺序是不对的呢？因为BST是左<中<右，所以我们可以用in-order traverse的方法：
In-order traverse模板：

public void inOrderTraverse{
    inOrderTraverse(root.left);
    //do something
    inOrderTraverse(root.right);
}

代码如下：

public class Solution {
    private TreeNode first = null, second = null;
    private TreeNode prev = new TreeNode(Integer.MIN_VALUE);
    
    public void recoverTree(TreeNode root) {
        Traverse(root);
        int temp = first.val;
        first.val = second.val;
        second.val = temp;
    }
    
    public void Traverse(TreeNode root) {
        if (root == null) return;
        
        Traverse(root.left);
        
        if (first == null && prev.val > root.val) {
            first = prev;
        }
        if (first != null && prev.val > root.val) {
            second = root;
        }
        prev = root;
        
        Traverse(root.right);
    }
}