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;
}
};
