二叉查找樹轉換爲有序雙向鏈表

將二叉排序樹轉換爲有序的雙向鏈表，只需調整結點的指針位置。見代碼

#include <stdio.h>
#include <stdlib.h>

#define TRUE 1
#define FALSE 0

typedef int Status;
typedef struct BSTNode
{
	int data;
	struct BSTNode *left, *right;
} BSTNode, *BSTree;

/*
*在二叉查找樹中查找數據域爲n個結點
*parent是當前結點的父結點，當T是根結點時，parent=NULL
*/
Status BSTSearch(BSTree T, int n, BSTree *parent)
{
	if (T == NULL)
		return FALSE;
	if (n == T->data)
	{
		*parent = T;
		return TRUE;
	}
	if (n < T->data)
	{
		*parent = T;
		return BSTSearch(T->left, n, parent);
	}
	if (n > T->data)
	{
		*parent = T;
		return BSTSearch(T->right, n, parent);
	}
}

/*
*在二叉查找樹中插入數據域爲n的結點
*/
Status BSTInsert(BSTree *T, int n)
{
	BSTree parent, s;
	parent = NULL;		//parent的初始值爲NULL
	if (BSTSearch(*T, n, &parent))
		return FALSE;
	s = (BSTree)malloc(sizeof(BSTNode));
	s->data = n;
	s->left = s->right = NULL;
	if (parent == NULL)
		*T = s;
	else if (n < parent->data)
		parent->left = s;
	else 
		parent->right = s;
	return TRUE;
}

/*
*中序遍歷二叉查找樹，結果是一個遞增數列
*/
void MidTraverse(BSTree T)
{
	if (T != NULL)
	{
		
		MidTraverse(T->left);
		printf("%d ", T->data);
		MidTraverse(T->right);
	}
}

/*
*將二叉排序樹轉換爲有序的雙向鏈表，只需調整結點的指針位置
*head指向有序雙向鏈表的第一個結點，tail指向最後一個結點
*/
void ToLinkedList(BSTree T, BSTree *head, BSTree *tail)
{
	BSTree lt, rh;
	if (T == NULL)
		*head = *tail = NULL;
	else
	{
		ToLinkedList(T->left, head, <);
		ToLinkedList(T->right, &rh, tail);
		if (lt)
		{
			lt->right = T;
			T->left = lt;
		}
		else 
			*head = T;
		if (rh)
		{
			T->right = rh;
			rh->left = T;
		}
		else
			*tail = T;
	}
}

int main(void)
{
	int i;
	int a[7] = {10, 6, 14, 4, 8, 16, 12};
	BSTree head, tail, p;
	BSTree T = NULL;
	for (i = 0; i < 7; i++)
		BSTInsert(&T, a[i]);
	MidTraverse(T);
	printf("\n");

	
	ToLinkedList(T, &head, &tail);
	p = head;
	while (p)		//向後遍歷有序雙向鏈表，結果是一個遞增序列
	{
		printf("%d ", p->data);
		p = p->right;
	}
	printf("\n");
	p = tail;
	while (p)		//向前遍歷有序雙向鏈表，結果是一個遞減序列
	{
		printf("%d ", p->data);
		p = p->left;
	}
	printf("\n");
	return 0;
}