LeetCode OJ

题目：

Two 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?

解题思路：

中序遍历BST，在遍历过程中记录出现错误的节点。err_1是第一次发生pre_val > root_val的节点，err_2是最后一次发生pre_val > root_val的节点。最后交换err_1->val 和 err_2->val

代码：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *err_1, *err_2;
    TreeNode *pre;
    
    void find2Nodes(TreeNode *root) {
        if (root == NULL) return;
        
        if (root->left != NULL) {
            find2Nodes(root->left);
        }
        if (pre != NULL && pre->val >= root->val) {
            if (err_1 == NULL) err_1 = pre;
            err_2 = root;
        }
        pre = root;
        if (root->right != NULL) {
            find2Nodes(root->right);
        }
    }
    
    void recoverTree(TreeNode *root) {
        if (root == NULL) return;
        
        err_1 = err_2 = pre = NULL;
        
        find2Nodes(root);
        swap(err_1->val, err_2->val);
    }
};