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);
    }
};