[LeetCode-75]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.
Analysis: The question requires O(1) space, let's first consider the case without this. How to traverse BST? Yep! The inOrder tree traversal.  (left->root->right) So the easiest way is inorder traverse the BST and find the element pair (two elements) which are not consistent with the definition of BST. In order to get the order, a queue is needed, which is O(n). Or we could just sort the result of the order and revalue the BST.
O(n) solution

void inorderBST(TreeNode *root, vector<int> &result, vector<TreeNode*> &list){
    if(root!=NULL){
        inorderBST(root->left,result,list);
        list.push_back(root);
        result.push_back(root->val);
        inorderBST(root->right,result,list);
    }
}
void recoverTree(TreeNode *root) {
    vector<int> result;
    vector<TreeNode*> nodelist;
    inorderBST(root,result,nodelist);
    sort(result.begin(),result.end());
    for(int i=0;i<nodelist.size();i++){
        nodelist[i]->val = result[i];
    }
}

Now how to do this procedure in O(1)?
What we need is actually two pointers, which point to 2 tree nodes where is incorrect. Therefore, we only need to store these two pointers, and, we also need another pointer to store the previous element, in order to  compare if the current element is valid or not.
In the code below, you will find, the last step is to replace the wrong pair's value. And the inOrder function is to search the whole BST and find the wrong pairs.
Note that, (1)the previous element is NOT the root node of the current element, but the previous element in the "inOrder"order; (2) To store the wrong pair, the first found wrong element is stored in first pointer, while the next is stored in the second pointer.
e.g. The correct BST is below:
The inorder traversal is :  1 3 4 6 7 8 10 13 14
If we change the value 4 and 8:  1 3 8 6 7 4 10 13 14, when it goes to the node 6, now the pre->val = 8, check if pre6). So we store the first pointer pointing to the node 6 and second pointer pointing to the node 8. To do so, we have stored the wrong nodes, if every other node keep the correct order, then swapping these nodes is enough for the problem. In other words, after the whole traversal, what we need to do is just changing the values of the first and second. Continue our example here, when the traversal goes to the node 4, now the node 7 is its pre, which is also wrong, so the second wrong node is found, and we change the second pointer pointing to node 4.

void getTwoNode(TreeNode *root,TreeNode *&n1, TreeNode*&n2, TreeNode*&pre){
    if(!root ) return;
    getTwoNode(root->left, n1, n2, pre);
    if(pre && pre->val > root->val){
        n2 = root;
        if(n1==NULL)
            n1 = pre;
        else n2 = root;
    }
    pre = root;
    getTwoNode(root->right,n1,n2,pre);
}
void recoverTree(TreeNode *root) {
    TreeNode *n1=NULL, *n2=NULL, *pre = NULL;
    if(!root) return;
    getTwoNode(root,n1,n2,pre);
   if(n1 && n2){
        int temp = n1->val;
        n1->val = n2->val;
        n2->val = temp;
    }

}

java

public class Solution {
    TreeNode r1;
	TreeNode r2;
	TreeNode pre;
	public void recoverTree(TreeNode root){
		r1 = null;
		r2 = null;
		pre = null;
		if(root == null) return;
		inorderBST(root);
		int temp = r1.val;
		r1.val = r2.val;
		r2.val = temp;
		return;
		
	}
	public void inorderBST(TreeNode root){
		if(root!=null){
			inorderBST(root.left);
			if(pre == null) pre = root;
			else {
				if(pre.val>root.val){
					if(r1 == null) r1 = pre;
					r2 = root;
				}
				pre = root;
			}
			inorderBST(root.right);
		}
	}
}