# Recover Binary Search Tree

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

Recover the tree without changing its structure.

**Example 1:**

```
Input: [1,3,null,null,2]

   1
  /
 3
  \
   2

Output: [3,1,null,null,2]

   3
  /
 1
  \
   2
```

**Example 2:**

```
Input: [3,1,4,null,null,2]

  3
 / \
1   4
   /
  2

Output: [2,1,4,null,null,3]

  2
 / \
1   4
   /
  3
```

**Follow up:**

* A solution using O(_n_) space is pretty straight forward.
* Could you devise a constant space solution?

```java
class Solution {
    public void recoverTree(TreeNode root) {
        TreeNode pre = null;
        TreeNode first = null, second = null;
        // Morris Traversal
        // IN ORDER traversal using Morris
        while (root != null) {
            if (root.left != null) {
                // connect threading for root
                TreeNode temp = root.left;
                while (temp.right != null && temp.right != root)
                    temp = temp.right;
                // the threading already exists then just deconstruct the thread
                if (temp.right != null) {
                    temp.right = null;

                    if (pre != null && pre.val > root.val) {
                        if (first == null)
                            first = pre;
                        second = root;
                    }
                    pre = root;
                    root = root.right;
                } else {
                    // construct the threading
                    temp.right = root;
                    root = root.left;
                }
            } else {
                if (pre != null && pre.val > root.val) {
                    if (first == null)
                        first = pre;
                    second = root;
                }
                pre = root;
                root = root.right;
            }
        }
        // swap two node values;
        if (first != null && second != null) {
            int t = first.val;
            first.val = second.val;
            second.val = t;
        }
    }
}
```
