(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");
}