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;

}