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In this post, you will find the solution for the **Recover Binary Search Tree** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either *Easy*, *Medium*, or *Hard*.

**LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:**

- Mathematics/Basic Logical Based Questions
- Arrays
- Strings
- Hash Table
- Dynamic Programming
- Stack & Queue
- Trees & Graphs
- Greedy Algorithms
- Breadth-First Search
- Depth-First Search
- Sorting & Searching
- BST (Binary Search Tree)
- Database
- Linked List
- Recursion, etc.

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** Link for the Problem** – Recover Binary Search Tree– LeetCode Problem

Recover Binary Search Tree– LeetCode Problem

**Problem:**

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]`

. `-2`

^{31}<= Node.val <= 2^{31}- 1

Recover Binary Search Tree– LeetCode Solutions

Recover Binary Search Tree Solution in C++:

class Solution { public: void recoverTree(TreeNode* root) { TreeNode* pred = nullptr; TreeNode* x = nullptr; // 1st wrong node TreeNode* y = nullptr; // 2nd wrong node stackstack; while (root || !stack.empty()) { while (root) { stack.push(root); root = root->left; } root = stack.top(), stack.pop(); if (pred && root->val < pred->val) { y = root; if (!x) x = pred; } pred = root; root = root->right; } swap(x, y); } void swap(TreeNode* x, TreeNode* y) { const int temp = x->val; x->val = y->val; y->val = temp; } };

Recover Binary Search Tree Solution in Java:

class Solution { public void recoverTree(TreeNode root) { TreeNode pred = null; TreeNode x = null; TreeNode y = null; Dequestack = new ArrayDeque<>(); while (root != null || !stack.isEmpty()) { while (root != null) { stack.add(root); root = root.left; } root = stack.pollLast(); if (pred != null && root.val < pred.val) { y = root; if (x == null) x = pred; } pred = root; root = root.right; } swap(x, y); } private void swap(TreeNode x, TreeNode y) { final int temp = x.val; x.val = y.val; y.val = temp; } }

Recover Binary Search Tree Solution in Python:

class Solution: def recoverTree(self, root: Optional[TreeNode]) -> None: pred = None x = None # 1st wrong node y = None # 2nd wrong node stack = [] while root or stack: while root: stack.append(root) root = root.left root = stack.pop() if pred and root.val < pred.val: y = root if not x: x = pred pred = root root = root.right def swap(x: Optional[TreeNode], y: Optional[TreeNode]) -> None: temp = x.val x.val = y.val y.val = temp swap(x, y)