99. Recover Binary Search Tree-优快云博客

本文链接：https://blog.youkuaiyun.com/nzcusing/article/details/58692747

本文介绍了一种在不改变二叉搜索树(BST)结构的情况下，修复因两个元素错误交换导致的问题的方法。通过中序遍历，找出并交换两个不符合升序规律的节点。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

问题描述
wo elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?
Subscribe to see which companies asked this question.
解决思路
使用inorder遍历的办法，因为在二叉查询树时，inorder遍历会满足遍历到的节点是有序的，而此时为升序，所以我们只要找到两个不满足pre->val < cur->val的节点就可以了。
代码

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