99. Recover Binary Search Tree

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

The solution to the above question is-

/**
 * 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* f=NULL;
    TreeNode* s=NULL;
    TreeNode* prev=new TreeNode (INT_MIN);
    
    void inorder(TreeNode* root){
        if(root==NULL)return;
        
        inorder(root->left);
        
        if(f==NULL && prev->val>root->val)f=prev;
        
        if(f!=NULL && prev->val > root->val)s=root;
        
        prev=root;
        
        inorder(root->right);
    }
    
    
    void recoverTree(TreeNode* root) {
        inorder(root);
        swap(f->val,s->val);
    }
};