程序员面试100题之一，把二元查找树变成排序的双向链表

#include "stdafx.h"

struct node
{
node * left;
node * right;
int value;
};

void convert(node * root, node *& last)
{
if(root == NULL)
return;

if(root->left)
convert(root->left, last);

root->left = last;

if(last)
last->right = root;

last = root;

if(root->right)
convert(root->right, last);

}

node * solution(node * root)
{
node * last = NULL;
convert(root, last);

while(last && last->left)
last = last->left;

return last;
}

int _tmain(int argc, _TCHAR* argv[])
{
node * n1 = new node();
node * n2 = new node();
node * n3 = new node();
node * n4 = new node();

n1->left = n2;
n1->right = n3;
n1->value = 3;

n2->left = n4;
n2->right = NULL;
n2->value = 2;

n3->left = NULL;
n3->right = NULL;
n3->value = 4;

n4->left = NULL;
n4->right = NULL;
n4->value = 1;

node * n = solution(n1);
while(n)
{
printf("%d ",n->value);
n = n->right;
}
return 0;
}