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].
  • -2^31 <= Node.val <= 2^31 - 1

Solution

Time complexity : O(n)
Space complexity : O(n)

/**
 * 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) {
        prev = first = second = nullptr;
        inorder(root);
        swap(first->val, second->val);
    }
    
    void inorder(TreeNode* node) {
        if (!node) return;
        
        inorder(node->left);
        if (prev && prev->val > node->val) {
            second = node;
            if (!first) first = prev;
        }
        prev = node;
        inorder(node->right);
    }
private:
    TreeNode *prev, *first, *second;
};

根據Binary Search Tree，在inorder traversal後會是已排序的性質下來做。
就找到prev比當前node大的做記錄。

跑完一輪inorder traversal後，找到兩個該交換的點，將其值交換即可。