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。

#include "stdlib.h"
#include "stdio.h"

typedef struct BSTNode
{
	char data;
	struct BSTNode *lnode;
	struct BSTNode *rnode;
}BSTNode;

BSTNode *CreateTree(char pre[], char in[], int l1, int r1, int l2, int r2){
	
	BSTNode *s;
	
	int i;
	if(l1 > r1)
		return NULL;

	s = (BSTNode *)malloc(sizeof(BSTNode));
	s->lnode = s->rnode = NULL;
	for(i=l2; i<=r2; ++i){
		if(pre[l1] == in[i]){
			break;
		}
	}
	s->data = in[i];

	s->lnode = CreateTree(pre, in, l1+1, l1+i-l2, l2, i-1);
	s->rnode = CreateTree(pre, in, l1+i-l2+1, r1, i+1, r2);

	return s;
}

void PrePrint(BSTNode *p){
	
	if(p != NULL){
		printf("%c ", p->data);
		
		PrePrint(p->lnode);
		PrePrint(p->rnode);
	}
}

void InPrint(BSTNode *p){
	
	if(p != NULL){
		InPrint(p->lnode);
		
		printf("%c ", p->data);
		
		InPrint(p->rnode);

	}
}

BSTNode *CreateDLNode(BSTNode *p, BSTNode *&prior){

	if(p != NULL){
		CreateDLNode(p->lnode, prior);
		if(prior != NULL)
			prior->rnode = p;
		p->lnode = prior;
		prior = p;
		CreateDLNode(p->rnode, prior);
	}

	return prior;
}

void main(){
	
	char pre[] = {'A','B','C','D','E','F','G'};
	char in[] = {'C','B','D','A','F','E','G'};

	BSTNode *p;
	p = CreateTree(pre, in, 0, 6, 0, 6);

	//InPrint(p);

	BSTNode *ss = NULL;

	CreateDLNode(p, ss);

	while(ss){
		printf("%c ", ss->data);
		ss = ss->lnode;
	}

}