二叉搜索树与双向链表

题目描述：
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。
关键点：1.递归
2.递归过程中链接左右指针

class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == NULL) return pRootOfTree;
        pRootOfTree = ConvertNode(pRootOfTree);
        while(pRootOfTree->left) pRootOfTree = pRootOfTree->left;
        return pRootOfTree;
    }
 
    TreeNode* ConvertNode(TreeNode* root)
    {
        if(root == NULL) return root;
        if(root->left)
        {
            TreeNode *left = ConvertNode(root->left);
            while(left->right) left = left->right;
            left->right = root;
            root->left = left;
        }
         
        if(root->right)
        {
            TreeNode *right = ConvertNode(root->right);
            while(right->left) right = right->left;
            right->left = root;
            root->right = right;
        }
        return root;
    }
};