题目描述

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?

• 思路
  二叉搜索树的两个节点发生了交换，恢复原来的结构。
  法一：定义两个全局变量记录交换的两个节点，还有一个表示前一个的节点。然后中序遍历判断大小（交换的节点分为相邻的和不相邻的），找到交换的两个节点，最后交换它们的值。参考博客
  法二：效率较低，中序遍历序列化此二叉树，然后排序。最后再中序遍历一遍赋对应的值。

//法一
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    TreeNode mis1 = null;
    TreeNode mis2 = null;
    TreeNode pre = null;
    
    public void recoverTree(TreeNode root) {
        if(root == null)
            return ;
        
        getMistake(root);
        
        if(mis1 != null && mis2 != null){
            int t = mis1.val;
            mis1.val = mis2.val;
            mis2.val = t;
        }
    }
    
    public void getMistake(TreeNode node){
        if(node == null)
            return ;
        
        if(node.left != null)
            getMistake(node.left);
        
        if(pre != null && node.val < pre.val){
            if(mis1 == null){
                mis1 = pre;
                mis2 = node; //交换的是相邻节点时
            }else{
                mis2 = node;
            }
        }
        pre = node;
        
        if(node.right != null)
            getMistake(node.right);
    }
}

//法二
import java.util.*;
public class Solution {
    public void recoverTree(TreeNode root) {
        if(root == null)
            return ;
        
        ArrayList<Integer> list = new ArrayList<Integer>();
        
        inOrder(root, list);
        
        Collections.sort(list);
        
        inOrderInit(root, list);
    }
    
    public void inOrderInit(TreeNode node, ArrayList<Integer> list){
        if(node != null){
            inOrderInit(node.left, list);
            node.val = list.get(0);
            list.remove(0);
            inOrderInit(node.right, list);
        }
    }
    
    public void inOrder(TreeNode node, ArrayList<Integer> list){
        if(node != null){
            inOrder(node.left, list);
            list.add(node.val);
            inOrder(node.right, list);
        }
    }
}