Problem Statement

You are given the root of a binary search tree (BST), where exactly two nodes of the tree were swapped by mistake. Recover the tree without changing its structure.

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

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

Problem Link

Recover Binary Search Tree - LeetCode

Code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode *first = NULL,*second = NULL,*p=NULL;
    
    void recoverTree(TreeNode* root) {
         
      **//  p = new TreeNode(INT_MIN);**
        inorder(root);
        int temp = first->val;
        first->val = second->val;
        second->val = temp;
        
    }
    void inorder(TreeNode* root)
    {
        if(!root)
        return;
    
        inorder(root->left);
    
        **if(!first&&(!p||p->val>=root->val)) 
        first = p;
    
        if(first&&p->val>=root->val)
        second = root;**
    
        p = root;
    
        inorder(root->right);
    
    
    }
};