二进制查找树转换为双向链表

全然依照海涛哥剑指offer里边的递归思路来写的。基本一样。仅作学习验证。努力锻炼。努力学习！

题目：输入一棵二元查找树，将该二元查找树转换成一个排序的双向链表。要求不能创建不论什么新的结点，仅仅调整指针的指向。

　　比方将二元查找树
                                            10
                                          /    \
                                        6       14
                                      /  \     /　 \
                                   　4     8  12 　  16
转换成双向链表

4=6=8=10=12=14=16。

code例如以下：

//Change a BSTree to a sorted double linklist
struct BSTreeNode 
{
	int value;
	BSTreeNode *left;
	BSTreeNode *right;
}head;
//Create a node of BSTree for test
BSTreeNode* CreateNode(int value)
{
	BSTreeNode *node = new BSTreeNode();
	node->value = value;
	node->left = NULL;
	node->right = NULL;
	return node;
}
//using a lastNode pointer to convert the pointer of the node
void ConvertNode(BSTreeNode *head, BSTreeNode **lastNode)
{
	if(head == NULL)
		return;
	BSTreeNode *pCurrent = head;
	if(pCurrent->left != NULL)
		ConvertNode(pCurrent->left, lastNode);
	pCurrent->left = *lastNode;
	if(*lastNode != NULL)
		(*lastNode)->right = pCurrent;
	*lastNode = pCurrent;
	if(pCurrent->right != NULL)
		ConvertNode(pCurrent->right, lastNode);
}
//lastNode pointer is pointing to the rear,so the head can be obtained 
BSTreeNode* Convert(BSTreeNode *head)
{
	if(head == NULL)
		return NULL;
	BSTreeNode *lastNode = NULL;
	ConvertNode(head, &lastNode);
	while(lastNode->left != NULL)
		lastNode = lastNode->left;
	return lastNode;
}
//main function for test
int main()
{
	BSTreeNode *head = CreateNode(10);
	head->left = CreateNode(8);
	head->left->left = CreateNode(7);
	head->left->right = CreateNode(9);
	head->right = CreateNode(12);
	head->right->left = CreateNode(11);
	head = Convert(head);
	while(head != NULL)
	{
		printf("%d ",head->value);
		head = head->right;
	}
}