Introduction

In this blog post, we will explore how to solve the LeetCode problem "99" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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?

Explanation

Here's the breakdown of the approach, complexities, and the Python code for recovering a swapped BST:

• Inorder Traversal: Perform an inorder traversal of the BST. This traversal should produce a sorted sequence if the tree is valid.
• Identify Swapped Nodes: During the inorder traversal, identify the two nodes that are out of order. The first node will be the first instance where prev.val > current.val. The second node will be the next instance where prev.val > current.val. If there is only one such instance, current node will be the second swapped node.

• Swap Values: Once the two nodes are identified, swap their values to restore the BST property.

• Time Complexity: O(N), where N is the number of nodes in the tree. Space Complexity: O(1)

Code

    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: TreeNode) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        self.first = None
        self.second = None
        self.prev = None

        def inorder(node):
            if not node:
                return

            inorder(node.left)

            if self.prev and self.prev.val > node.val:
                if not self.first:
                    self.first = self.prev
                    self.second = node  # Initialize second in case only adjacent nodes are swapped
                else:
                    self.second = node

            self.prev = node
            inorder(node.right)

        inorder(root)

        # Swap the values of the two identified nodes.
        self.first.val, self.second.val = self.second.val, self.first.val