【LeetCode 99】Recover Binary Search Tree

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

题意：

　　一颗二叉搜索树中有2个结点的元素被误换了，要求恢复二叉搜索树的原状。

思路：

　　中序遍历BST，若发现一次逆序，说明是两个相连结点的误换，直接交换结点的值；若发现两次逆序，则为不相连的两个结点误换，交换错误的结点即可。用一个变量保存先前结点的状态，若发生逆序则记录，空间复杂度为常量。

C++：

 /**
  * 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:

     vector<TreeNode* > errs;
     TreeNode *preNode;

     void rec(TreeNode *root)
     {
         if(root->left != )
             rec(root->left);

         if(preNode !=  && root->val < preNode->val)
         {
             errs.push_back(preNode);
             errs.push_back(root);
         }

         preNode = root;

         if(root->right != )
             rec(root->right);
     }

     void recoverTree(TreeNode* root) {
         if(root == )
             return ;

         preNode = ;

         rec(root);

         int temp = , index = ;

         if(errs.size() == )
             index = ;

         temp = errs[]->val;
         errs[]->val = errs[index]->val;
         errs[index]->val = temp;
     }
 };