[LeetCode] Recover Binary Search Tree

wo 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.

SOLUTION: use inorder traversal to check whether every node is in increasing order. Keep two pointers to the nodes out of order.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:

    //return the previous visited node
    TreeNode* inorder(TreeNode *root,TreeNode *prev, TreeNode*& n1, TreeNode*& n2)
    {
        if(root==NULL) return prev;
        
        prev=inorder(root->left,prev,n1,n2);
        
        //check whether the root is out of order
        if(prev!=NULL && root->val < prev->val)
        {
            if(n1==NULL)
                n1=prev;
            //swap nodes
            n2=root;
        }
        
        prev=root;
        return inorder(root->right,prev,n1,n2);
    }

    void recoverTree(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        TreeNode *n1=NULL;
        TreeNode *n2=NULL;
        TreeNode *parent=NULL;
        inorder(root,parent,n1,n2);
        if(n1!=NULL && n2!=NULL)
        {
        int tmp=n1->val;
        n1->val=n2->val;
        n2->val=tmp;
        }
    }
};