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Recover Binary Search Tree LeetCode Solution

Problem – Recover Binary Search Tree LeetCode Solution

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.


  • 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?

Recover Binary Search Tree LeetCode Solution in Python

class TreeNode:
    def __init__(self, val, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
    def __repr__(self):
        return 'TreeNode({})'.format(self.val)
def deserialize(string):
    if string == '{}':
        return None
    nodes = [None if val == 'null' else TreeNode(int(val))
             for val in string.strip('[]{}').split(',')]
    kids = nodes[::-1]
    root = kids.pop()
    for node in nodes:
        if node:
            if kids: node.left  = kids.pop()
            if kids: node.right = kids.pop()
    return root

def drawtree(root):
    def height(root):
        return 1 + max(height(root.left), height(root.right)) if root else -1
    def jumpto(x, y):
        t.goto(x, y)
    def draw(node, x, y, dx):
        if node:
            t.goto(x, y)
            jumpto(x, y-20)
            t.write(node.val, align='center', font=('Arial', 12, 'normal'))
            draw(node.left, x-dx, y-60, dx/2)
            jumpto(x, y-20)
            draw(node.right, x+dx, y-60, dx/2)
    import turtle
    t = turtle.Turtle()
    t.speed(0); turtle.delay(0)
    h = height(root)
    jumpto(0, 30*h)
    draw(root, 0, 30*h, 40*h)
if __name__ == '__main__':

Recover Binary Search Tree LeetCode Solution in C++

class Solution {
	TreeNode* firstMistake, *secondMistake, *pre;
	void recoverTree(TreeNode* root) {
		pre = new TreeNode(INT_MIN);
		swap(firstMistake->val, secondMistake->val);

	void inorder(TreeNode* root) {
		if(root == nullptr) 


		if(firstMistake == nullptr && root->val < pre->val)
			firstMistake = pre;
		if(firstMistake != nullptr && root->val < pre->val)
			secondMistake = root;
		pre = root;


Recover Binary Search Tree LeetCode Solution in Java

class Solution {
    TreeNode prev = null, first = null, second = null;

    public void recoverTree(TreeNode root) {
        int temp = first.val;
        first.val = second.val;
        second.val = temp;

    private void evalSwappedNodes(TreeNode curr) {
        if (curr == null)
        if (prev != null && prev.val > curr.val) {
            if (first == null)
                first = prev;
            second = curr;
        prev = curr;
Recover Binary Search Tree LeetCode Solution Review:

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