Recover Binary Search Tree

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?

 

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1

  / \

 2   3

    /

   4

    \

     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

修正二叉树中错误的两个结点，在不改变结构的情况下。

思路：首先想到的就是中序遍历二叉搜索树，中序遍历二叉搜索树的结点是按从小到大的顺序，如果出现该节点值比前一个结点小，则说明这个值不合法。用pPre存中序遍历的前一个结点，方便比较大小，first和second分别保存较大的和较小的值。

/**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

class Solution {

public:

    void recoverMistake(TreeNode *root,TreeNode *&first,TreeNode *&second,TreeNode *&pPre)

    {

        if(root==NULL)

            return;

        recoverMistake(root->left,first,second,pPre);

        if(pPre && pPre->val>root->val)

        {

            if(first==NULL)

            {

                first=pPre;

                second=root;

            }

            else

            {

                second=root;

            }

        }

        pPre=root;

        recoverMistake(root->right,first,second,pPre);

    }

    void recoverTree(TreeNode *root) {

        TreeNode *first=NULL,*second=NULL,*pPre=NULL;

        recoverMistake(root,first,second,pPre);

        swap(first->val,second->val);

    }

};