LeetCode-99. Recover Binary Search TreeLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/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].
• -2^31 <= Node.val <= 2^31 - 1

Follow up: A solution using O(n) space is pretty straight-forward. Could you devise a constant O(1) space solution?

【C++】

/**
 * 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:
    void recoverTree(TreeNode* root) {
        TreeNode *mistake1 = nullptr, *mistake2 = nullptr;
        TreeNode *prev = new TreeNode(INT_MIN);
        inorder(root, mistake1, mistake2, prev);
        if (mistake1 && mistake2) {
            int temp = mistake1->val;
            mistake1->val = mistake2->val;
            mistake2->val = temp;
        }
    }
    
    void inorder(TreeNode* root,
                 TreeNode*& mistake1,
                 TreeNode*& mistake2,
                 TreeNode*& prev) {
        if (root == nullptr) {return;}
        if (root->left) {inorder(root->left, mistake1, mistake2, prev);}
        if (prev && root->val < prev->val) {
            if (mistake1 == nullptr) {
                mistake1 = prev;
                mistake2 = root;
            } else {mistake2 = root;}
        }
        prev = root;
        if (root->right) {inorder(root->right, mistake1, mistake2, prev);}
    }
};

【Java】

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private TreeNode mistake1 = null;
    private TreeNode mistake2 = null;
    private TreeNode prev = new TreeNode(Integer.MIN_VALUE);
    
    public void recoverTree(TreeNode root) {
        inorder(root);
        if (mistake1 != null && mistake2 != null) {
            int temp = mistake1.val;
            mistake1.val = mistake2.val;
            mistake2.val = temp;
        }
    }
    
    void inorder(TreeNode root) {
        if (root == null) {return;}
        if (root.left != null) {inorder(root.left);}
        if (root.val < prev.val) {
            if (mistake1 == null) {
                mistake1 = prev;
                mistake2 = root;
            } else {
                mistake2 = root;
            }
        }
        prev = root;
        if (root.right != null) {
            inorder(root.right);
        }
    }
}