題目：

Two elements of a binary search tree (BST) are swapped by mistake. 

Recover the tree without changing its structure. 

思路：

中序遍歷找到發生錯誤的兩個節點，並記錄下節點，最後交換這兩個節點的值即可。pre代表前序遍歷的前一個節點，mistake1和mistake2是兩個錯誤的節點。判斷出錯的依據是當前節點的值大於pre節點的值。

程式碼：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
        TreeNode pre,mistake1,mistake2;
    public void recoverTree(TreeNode root) {
        inTraversal(root);
        int 
 
temp=mistake1.val;
        mistake1.val = mistake2.val;
        mistake2.val = temp;
    }
    public void inTraversal(TreeNode root){
        if(root==null) return ;
        inTraversal(root.left);
        if(pre!=null&&pre.val>root.val){
            if(mistake1==null){
                mistake1
 
=pre;
                mistake2=root;
            }else mistake2=root;
        }
        pre=root;
        inTraversal(root.right);
    }
}