# 99. Recover Binary Search Tree

#### Medium

***

You are given the `root` of a binary search tree (BST), where the values of **exactly** two nodes of the tree were swapped by mistake. *Recover the tree without changing its structure*.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/10/28/recover1.jpg)

```
Input: root = [1,3,null,null,2]
Output: [3,1,null,null,2]
Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid.
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/10/28/recover2.jpg)

```
Input: root = [3,1,4,null,null,2]
Output: [2,1,4,null,null,3]
Explanation: 2 cannot be in the right subtree of 3 because 2 < 3. Swapping 2 and 3 makes the BST valid.
```

&#x20;

**Constraints:**

* The number of nodes in the tree is in the range `[2, 1000]`.
* `-231 <= Node.val <= 231 - 1`

&#x20;

**Follow up:** A solution using `O(n)` space is pretty straight-forward. Could you devise a constant `O(1)` space solution?

#### Solution 1 : Using Recursion

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def recoverTree(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        self.first = None
        self.second = None
        self.prev = TreeNode(-float("inf"))
        self.traverse(root)
        self.first.val, self.second.val = self.second.val, self.first.val
    
    def traverse(self, root):
        if not root:
            return
        self.traverse(root.left)
        if self.first is None and self.prev.val >= root.val:
            self.first = self.prev
        if self.first is not None and self.prev.val >= root.val:
            self.second = root
        self.prev = root
        self.traverse(root.right)

```

#### Solution 2 : Morris Traversal

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def recoverTree(self, root: Optional[TreeNode]) -> None:
        # Morris Traversal
        # If you get confused , especially with condition at Line 22, read
        # https://stackoverflow.com/questions/5502916/explain-morris-inorder-tree-traversal-without-using-stacks-or-recursion
        current, arr = root, []
        prev = TreeNode(-float("inf"))
        while current:
            if current.left is None:
                if prev.val > current.val:
                    arr += [prev, current]
                prev = current
                current = current.right
            else:
                pre = current.left
                while pre.right and pre.right != current:
                    pre = pre.right
                if pre.right is None:
                    pre.right = current
                    current = current.left
                else:
                    pre.right = None
                    if prev.val > current.val:
                        arr += [prev, current]
                    prev = current
                    current = current.right
        arr[0].val, arr[-1].val = arr[-1].val, arr[0].val
 
```