题目：

输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向

 

代码:

思路采用递归的方式实现（类似树的中序遍历），定义一个已经调整好的链表的左端节点指针和右端节点指针

#include "pch.h"
#include <iostream>
#include <sstream>
#include <string>

using namespace std;

struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};
class Solution {
public:
	TreeNode *L = NULL, *R = NULL;
	TreeNode* Convert(TreeNode* pRootOfTree)
	{
		change(pRootOfTree);
		return L;
	}
	void change(TreeNode* pRoot) {
		if (pRoot->right == NULL && pRoot->left == NULL) {
			if (L == NULL && R == NULL)
				L = pRoot, R = pRoot;
			else {
				R->right = pRoot;
				pRoot->left = R;
				R = pRoot;
			}
			return;
		}
		else {
			if (pRoot->left)change(pRoot->left);
			if (L == NULL && R == NULL)L = pRoot, R = pRoot;
			else {
				R->right = pRoot;
				pRoot->left = R;
				R = pRoot;
			}
			if (pRoot->right)change(pRoot->right);
		}
	}
};

//用来进行测试的代码

//将输入的string转换为一种任意已知类型的函数
template <class T>
T string_to_double(const string& str)
{
	istringstream iss(str);
	T dou_num;
	iss >> dou_num;
	return dou_num;
}

void creat_tree(TreeNode* &R)
{
	//int r;
	string str;
	cin >> str;
	if (str == "#")
	{
		R = NULL;
	}
	else
	{
		int r = string_to_double<int>(str);
		TreeNode *root = new TreeNode(r);
		cout << "root(" << r << ")------>Leftchildren:" << endl;
		creat_tree(root->left);
		cout << "root(" << r << ")------>rightchildren:" << endl;
		creat_tree(root->right);
		R = root;
	}
}
void main() 
{
	/*
	TreeNode *r = new TreeNode(2);
	r->left = new TreeNode(1);
	r->right = new TreeNode(3);
	*/
	TreeNode *r;
	cout << "请输入根节点:" << endl;
	creat_tree(r);
	Solution s;
	s.Convert(r);
	TreeNode *ss = s.L;
	while (ss != s.R) {
		cout << ss->val << " ";
		ss = ss->right;
	}
	cout << s.R->val << endl;
	ss = s.R;
	while (ss != s.L) {
		cout << ss->val << " ";
		ss = ss->left;
	}
	cout << s.L->val << endl;
	cout <<"result"<< endl;
}

结果：

注：在这个题上有一些问题，下面的代码在vs2017中能够通过，写了几个例子都没有问题，但是在牛客网上编译结果总是有段错误，没有找到原因