Leetcode-Recover BST

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?

O(n) time and O(1) space solution: Morris Traversal

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
    


public class Solution {
    public void recoverTree(TreeNode root) {
        if (root==null) return;
        
        TreeNode pre=null,cur=null,first=null,second=null;
        cur = root;
        while (cur!=null){
            //cur.left is null.
            if (cur.left==null){
                if (pre!=null && pre.val>cur.val){
                    if (first==null){
                        first = pre;
                        second = cur;
                    } else second = cur;
                }
                pre = cur;
                cur = cur.right;
            } else {
                //get predecessor.
                TreeNode temp = getPredecessor(cur);
                if (temp.right==null){
                    temp.right=cur;
                    cur = cur.left;
                } else {
                    if (pre!=null && pre.val>cur.val){
                        if (first==null){
                            first = pre;
                            second = cur;
                        } else second = cur;
                    }
                    temp.right = null;
                    pre = cur;
                    cur = cur.right;
                }
            }
        }
        
        if (first==null) return;
        
        int temp = first.val;
        first.val = second.val;
        second.val = temp;
        
        return;
    }
    
    public TreeNode getPredecessor(TreeNode cur){
        TreeNode pre = cur.left;
        while (pre.right!=null && pre.right!=cur){
            pre = pre.right;
        }
        
        return pre;
    }
}

O(log(n)) space solution:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Result {
    TreeNode pre;
    TreeNode first;
    TreeNode second;
    Result() {
        pre = first = second = null;
    }
}

public class Solution {
    public void recoverTree(TreeNode root) {
        Result res = new Result();
        recoverTreeRecur(root,res);
        if (res.first!=null && res.second!=null){
            int temp = res.first.val;
            res.first.val = res.second.val;
            res.second.val = temp;
        }
    }

    public void recoverTreeRecur(TreeNode cur, Result res){
        if (cur==null)
            return;

        recoverTreeRecur(cur.left, res);
        if (res.pre==null) res.pre = cur;
        else if (res.pre.val>cur.val){
            if (res.first==null)
                res.first = res.pre;
            res.second = cur;
        }

        res.pre = cur;
        recoverTreeRecur(cur.right,res);
    }        
}