輸入一顆二叉搜尋樹，將該二叉搜尋樹轉換成一個排序的雙向連結串列。

解題思路

  這個題目可以使用分治法完成，C++版程式碼如下：

#include <iostream>
#include <algorithm>
#include <math.h>
#include <cstring>
#include "ListNode.h"
#include "TreeNode.h"
#include "Graph.h"
using namespace std;

#define MAXNUM 100010
#define DRIFT 1001

void ConvertNode(TreeNode* pRoot, TreeNode** pLastNodeInList){
    if(pRoot == nullptr)
        return ;
    TreeNode* pCurrent = pRoot;

    if(pCurrent->left != nullptr)
        ConvertNode(pCurrent->left, pLastNodeInList);

    pCurrent->left = *pLastNodeInList;
    if(*pLastNodeInList != nullptr)
        (*pLastNodeInList)->right = pCurrent;
    *pLastNodeInList = pCurrent;

    if(pCurrent->right != nullptr)
        ConvertNode(pCurrent->right, pLastNodeInList);
}

TreeNode* Convert(TreeNode* pRootOfTree){
    TreeNode* pLastNodeInList = nullptr;
    ConvertNode(pRootOfTree, &pLastNodeInList);

    // pLastNodeInList指向雙向連結串列的尾結點
    // 我們需要返回頭結點
    TreeNode* pHeadOfList = pLastNodeInList;
    while(pHeadOfList != nullptr && pHeadOfList->left != nullptr)
        pHeadOfList = pHeadOfList->left;

    return pHeadOfList;
}

int main()
{
    return 0;
}