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.

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.

/**

 * 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 inOrder(TreeNode *&pre,TreeNode *&mis1,TreeNode *&mis2, TreeNode *root)

{

    if(root==NULL)return;

    inOrder(pre,mis1,mis2,root->left);

    if(pre!=NULL&&pre->val>root->val)

    {

        if(mis1==NULL)

        {

            mis1=pre;

            mis2=root;

        }

        else

        {

            mis2=root;

        }

    }

    pre=root;

    inOrder(pre,mis1,mis2,root->right);

}

void recoverTree(TreeNode *root) 

{

    TreeNode *pre=NULL, *mis1=NULL, *mis2=NULL;

    inOrder(pre,mis1,mis2,root);

    if(mis1&&mis2)

    {

        int tmp=mis1->val;

        mis1->val=mis2->val;

        mis2->val=tmp;

    }

}

/*

private:

   TreeNode *pre=NULL, *mis1=NULL, *mis2=NULL;

public:

void inOrder(TreeNode *root)

{

    if(root==NULL)return;

    inOrder(root->left);

    if(pre!=NULL&&pre->val>root->val)

    {

        if(mis1==NULL)

        {

            mis1=pre;

            mis2=root;

        }

        else

        {

            mis2=root;

        }

    }

    pre=root;

    inOrder(root->right);

}

    void recoverTree(TreeNode *root) {

        inOrder(root);

        if(mis1&&mis2)

        {

            int tmp=mis1->val;

            mis1->val=mis2->val;

            mis2->val=tmp;

        }

    }

*/

};