# Question

Formatted question description: https://leetcode.ca/all/99.html

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: 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: 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.


Constraints:

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

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

# Algorithm

Three pointers are needed, first and second respectively represent the first and second disordered nodes, and pre points to the previous node of the current node’s in-order traversal.

The traditional in-order traversal recursion is used here, but where the node value should be output, it is replaced by judging the size of pre and the current node value. If pre is larger, if first is empty, first point to the node pointed to by pre, or else, set second to the current node.

In this way, the in-order traverses the entire tree, and if both first and second exist, then their node values can be exchanged. The space complexity of this algorithm is still O(n), where n is the height of the tree.

# Code

• 
public class Recover_Binary_Search_Tree {

public static void main(String[] args) {

}

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/

public class Solution {

TreeNode n1;
TreeNode n2;
TreeNode prev; // @note:@memorize: key is use prev!

public void recoverTree(TreeNode root) {
inOrderTraversal(root); // will record 2 swapped nodes
swap(); // swap 2 nodes
}

private void inOrderTraversal(TreeNode root) {

if (root == null) {
return;
}

inOrderTraversal(root.left);

// @note: 1,6,3,4,5,2 => NOT next to each other => n1=6-prev, but n2=2-root
// @note: 1,2,4,3,5,6 => when next to each other, n2 needs to be root
// bigger than next, then wrong

if (prev != null && prev.val >= root.val) {
if (n1 == null) {
n1 = prev;
n2 = root; // @note:@memorize: here is key, for case when in total two nodes
} else {
// n2 = prev; @note:@memorize: n1 and n2, n1=prev but n2=root
n2 = root;
}
}

prev = root;

inOrderTraversal(root.right);
}

// @note: concern about n1 or n2 is null? by question definition, if swapped, then at least 2 nodes
private void swap() {
int tmp = n1.val;
n1.val = n2.val;
n2.val = tmp;
}
}

}

############

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private TreeNode prev;
private TreeNode first;
private TreeNode second;

public void recoverTree(TreeNode root) {
dfs(root);
int t = first.val;
first.val = second.val;
second.val = t;
}

private void dfs(TreeNode root) {
if (root == null) {
return;
}
dfs(root.left);
if (prev != null && prev.val > root.val) {
if (first == null) {
first = prev;
}
second = root;
}
prev = root;
dfs(root.right);
}
}

• // OJ: https://leetcode.com/problems/recover-binary-search-tree/
// Time: O(N)
// Space: O(H)
class Solution {
TreeNode *a = NULL, *b = NULL;
void update(TreeNode *x, TreeNode *y) {
if (!a || x->val < a->val) a = x;
if (!b || y->val > b->val) b = y;
}
void dfs(TreeNode *root, TreeNode *left = NULL, TreeNode *right = NULL) {
if (!root) return;
if (left && left->val > root->val) update(root, left);
if (right && right->val < root->val) update(right, root);
dfs(root->left, left, root);
dfs(root->right, root, right);
}
public:
void recoverTree(TreeNode* root) {
dfs(root);
swap(a->val, b->val);
}
};

• # 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.
"""

def dfs(root):
if root is None:
return
# inside a dfs(), 'nonlocal' is must if 'prev' being modified
nonlocal prev, first, second
dfs(root.left)
if prev and prev.val > root.val:
if first is None:
first = prev
second = root
prev = root
dfs(root.right)

prev = first = second = None
dfs(root)
first.val, second.val = second.val, first.val

############

class Solution:
def __init__(self):
self.n1 = None
self.n2 = None
self.pre = None

if root is None: return
if self.pre is not None:
if root.val < self.pre.val:
if self.n1 is None:
self.n1 = self.pre
self.n2 = root
else:
self.n2 = root
self.pre = root

def recoverTree(self, root):
if self.n1 is not None and self.n2 is not None:
self.n1.val, self.n2.val = self.n2.val, self.n1.val


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func recoverTree(root *TreeNode) {
var prev, first, second *TreeNode
var dfs func(*TreeNode)
dfs = func(root *TreeNode) {
if root == nil {
return
}
dfs(root.Left)
if prev != nil && prev.Val > root.Val {
if first == nil {
first = prev
}
second = root
}
prev = root
dfs(root.Right)
}
dfs(root)
first.Val, second.Val = second.Val, first.Val
}

• /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {void} Do not return anything, modify root in-place instead.
*/
var recoverTree = function (root) {
let prev = null;
let first = null;
let second = null;
function dfs(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);
const t = first.val;
first.val = second.val;
second.val = t;
};


• /**
* Definition for a binary tree node.
* public class TreeNode {
*     public int val;
*     public TreeNode left;
*     public TreeNode right;
*     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
public class Solution {
private TreeNode prev, first, second;

public void RecoverTree(TreeNode root) {
dfs(root);
int t = first.val;
first.val = second.val;
second.val = t;
}

private void dfs(TreeNode root) {
if (root == null) {
return;
}
dfs(root.left);
if (prev != null && prev.val > root.val) {
if (first == null) {
first = prev;
}
second = root;
}
prev = root;
dfs(root.right);
}
}