二叉搜索树与双向链表 【第一题】

#include <iostream>

using namespace std;

typedef struct BinaryTree
{
	struct BinaryTree *left,*right;
	int data;
}BinaryTree;

void initTree(BinaryTree **p);
void printList(BinaryTree *list);
BinaryTree *treeToList(BinaryTree *pTree);

int main(void)
{
	BinaryTree *pTree = NULL,*pList = NULL;

	initTree(&pTree);
	pList = treeToList(pTree);
	printList(pList);
	return 0;
}
//		1
//	   / //    2   3
//	 / //	4   5
void initTree(BinaryTree **p)
{
	*p = new BinaryTree;
	(*p)->data = 1;

	BinaryTree *tmpNode = new BinaryTree;
	tmpNode->data = 2;
	(*p)->left = tmpNode;

	tmpNode = new BinaryTree;
	tmpNode->data = 3;
	(*p)->right = tmpNode;
	tmpNode->left = NULL;
	tmpNode->right = NULL;

	BinaryTree *currentNode = (*p)->left;

	tmpNode = new BinaryTree;
	tmpNode->data = 4;
	currentNode->left = tmpNode;
	tmpNode->left = NULL;
	tmpNode->right = NULL;

	tmpNode = new BinaryTree;
	tmpNode->data = 5;
	currentNode->right = tmpNode;
	tmpNode->left = NULL;
	tmpNode->right = NULL;

}
void printList(BinaryTree *list)
{
	while(list!=NULL)
	{
		cout << list->data;
		if(list->right!=NULL)
		{
			cout << "<==>";
		}
		list = list->right;
	}
}
//---------------核心代码-----------------------
void convert(BinaryTree *pTree,BinaryTree **pre)
{
	if(pTree == NULL)
		return;
	BinaryTree *pCurrent = pTree;
	if(pCurrent->left != NULL)
		convert(pTree->left,pre);

	//当前点的前驱为pre
	pCurrent->left = *pre;
	//pre不为NULL，pre的后继为当前点
	if(*pre != NULL)
		(*pre)->right = pCurrent;
	//pre为当前点
	*pre = pCurrent;

	if(pCurrent->right != NULL)
		convert(pTree->right,pre);
}
BinaryTree *treeToList(BinaryTree *pTree)
{
	BinaryTree *pre = NULL;

	convert(pTree,&pre);

	//按照初始化的树，链表为4<==>2<==>5<==>1<==>3
	//此时pre在结点3处，应该返回到结点4处，再遍历输出
	while(pre!=NULL && pre->left!=NULL)
		pre = pre->left;

	return pre;
}