Recover Binary Search Tree

Recover Binary Search Tree

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

Solutions

🧠 Cpp

/**
 * 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
{
    TreeNode* find_X_then_value(TreeNode *root, int val, function<bool(int,int)> X)
    {
        if(!root)
            return nullptr;

        int val_to_check = val;
        if(X(root->val, val))
            val_to_check = root->val;


        TreeNode *res_left = find_X_then_value(root->left, val_to_check, X),
                 *res_right = find_X_then_value(root->right, val_to_check, X);

        if(res_left && res_right)
            return X(res_left->val, res_right->val) ? res_left :res_right;
        else if(res_left)
            return res_left;
        else if(res_right)
           return res_right; 
        else if(X(root->val, val))
            return root;


        return nullptr;

    }

public:
    void recoverTree(TreeNode* root)
    {
        if(!root)
            return;

        TreeNode *misplaced_left = find_X_then_value(root->left, root->val, std::greater<int>()),
                 *misplaced_right =  find_X_then_value(root->right, root->val, std::less<int>());

        if(misplaced_left && misplaced_right)
            std::swap(misplaced_left->val, misplaced_right->val);
        else if(misplaced_left)
            std::swap(root->val, misplaced_left->val);
        else if(misplaced_right)
            std::swap(root->val, misplaced_right->val);
        else
        {
            recoverTree(root->left);
            recoverTree(root->right);
        }

    }
};