99. 恢复二叉搜索树

二叉搜索树中的两个节点被错误地交换。

请在不改变其结构的情况下，恢复这棵树。

示例 1:

输入: [1,3,null,null,2]

1
/
3

2

输出: [3,1,null,null,2]

3
/
1

2
示例 2:

输入: [3,1,4,null,null,2]

3
/
1 4
/
2

输出: [2,1,4,null,null,3]

2
/
1 4
/
3
进阶:

使用 O(n) 空间复杂度的解法很容易实现。
你能想出一个只使用常数空间的解决方案吗？

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/recover-binary-search-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
     
    public void recoverTree(TreeNode root) {
     
        TreeNode x = null, y = null, pred = null, predecessor = null;
        while(root != null){
     
            if(root.left != null){
     //左子树不为空，就找左子树最右边的结点
                predecessor = root.left;
                while(predecessor.right != null && predecessor.right != root){
     
                    predecessor = predecessor.right;
                }
                if(predecessor.right == null){
     
                    predecessor.right = root;
                    root = root.left;
                }
                else{
     
                    if(pred != null && root.val < pred.val){
     
                        y = root;
                        if(x == null){
     
                            x = pred; 
                        }
                    }
                    pred = root;
                    predecessor.right = null;
                    root = root.right;
                }
            }
            else{
     
                if(pred != null && root.val < pred.val){
     
                    y = root;
                    if(x == null){
     
                        x = pred;
                    }
                }
                pred = root;
                root = root.right;
            }
        }
        swap(x,y);
    }
     public void swap(TreeNode x, TreeNode y) {
     
        int tmp = x.val;
        x.val = y.val;
        y.val = tmp;
    }

}