Recover Binary Search Tree，恢复二叉排序树

问题描述：题意就是二叉树中有两个节点交换了，恢复结构。

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

Recover the tree without changing its structure.

算法分析：其实还是利用中序遍历，每次记录前一个访问的节点，和当前节点就行比较，如果逆序则记录下来。两个节点交换有两种情况，一种是相邻节点交换，那么只有一个逆序，非相邻节点交换，则有两个逆序。

public class RecoverBinarySearchTree 
{
    TreeNode mistake1 = null;
    TreeNode mistake2 = null;
    TreeNode pre = null;
    public void recoverTree(TreeNode root) 
    {
        inorderTraverse(root);
        int temp = mistake1.val;
        mistake1.val = mistake2.val;
        mistake2.val = temp;
    }
    public void inorderTraverse(TreeNode root)
    {
        if(root == null)
        {
            return;
        }
        inorderTraverse(root.left);
        if(pre != null)
        {
            if(pre.val >= root.val)//将当前要访问的根节点和pre比较
            {
                if(mistake1 == null)
                {
                    mistake1 = pre;
                }
                mistake2 = root;
            }
        }
        pre = root;//访问完根节点时，将根节点置为pre，相当于pre指针后移一位
        inorderTraverse(root.right);
    }
}