(微软100题)1.把二元查找树转变成排序的双向链表

#include <iostream>
using namespace std;

/*
1.把二元查找树转变成排序的双向链表
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。
要求不能创建任何新的结点，只调整指针的指向。

10
/  
6	14
/ 	  / 
4  8 12 16

转换成双向链表
4=6=8=10=12=14=16。
*/

struct Node{
	int data;
	Node *left;
	Node *right;
	Node(int d = 0, Node *lr = 0, Node *rr = 0):
	data(d), left(lr), right(rr)
	{
	}
};

Node *create()
{
	Node *root;
	Node *p4 = new Node(4);
	Node *p8 = new Node(8);
	Node *p6 = new Node(6, p4, p8);

	Node *p12 = new Node(12);
	Node *p16 = new Node(16);
	Node *p14 = new Node(14, p12, p16);

	Node *p10 = new Node(10, p6, p14);
	root = p10;

	return root;
}

Node *change(Node *p, bool asRight)
{
	if (NULL == p)
		return NULL;
	Node *pLeft = change(p->left, false);
	if (pLeft)
		pLeft->right = p;
	p->left = pLeft;

	Node *pRight = change(p->right, true);
	if (pRight)
		pRight->left = p;
	p->right = pRight;

	Node *r = p;
	if (asRight)
	{
		while (r->left)
			r = r->left;
	}else
	{
		while (r->right)
			r = r->right;
	}
	return r;
}

void main(){
	Node *root = create();
	Node *tail = change(root, false);
	while (tail)
	{
		cout << tail->data << " ";
		tail = tail->left;
	}
	cout << endl;

	root = create();
	Node *head = change(root, true);
	while (head)
	{
		cout << head->data << " ";
		head = head->right;
	}
	cout << endl;
	system("pause");
}