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。

//递归解法
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
class Solution {
public:

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

        TreeNode* pHeadOfList=pLastNodeInList;
        while(pHeadOfList->left!=nullptr)
        {
            pHeadOfList=pHeadOfList->left;
        }
        return pHeadOfList;
    }

    //将子树进行中序遍历并转换为双向链表
    void ConvertNode(TreeNode* pNode,TreeNode* &pLastNodeInList)
    {
        if(pNode==nullptr)
            return;

        TreeNode* pCurrent=pNode;

        //若左子树不为空，则将左子树转换为双向链表并且改变当前节点pNode的指向
        if(pCurrent->left)
            ConvertNode(pCurrent->left,pLastNodeInList);
        pCurrent->left=pLastNodeInList;
        if(pLastNodeInList)
            pLastNodeInList->right=pCurrent;
        //将左子树转换完之后，双向链表最后一个指针是当前节点
        pLastNodeInList=pNode;

        //若右子树不为空，则按一样步骤处理右子树
        if(pCurrent->right)
            ConvertNode(pCurrent->right,pLastNodeInList);
    }
};