Recover Binary Search Tree LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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 is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

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.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

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].
• -231 <= Node.val <= 231 - 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

    stack<TreeNode*> stack;

    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;

    Deque<TreeNode> stack = 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)