6.3.2 层次遍历

输入一颗二叉树，你的任务是按从上到下，从左到右的顺序输出各个节点的值。每个节点都按照从根节点到它的移动序列给出（L表示左，R表示右）。在输入中，每个节点的左括号之间没有空格，相邻节点之间用一个空格隔开。每棵树的输入用一对空括号（）结束（这对括号本身不代表一个结点），如图所示。

                                                                                      5

                                                                           4                     8

                                                                    11                   13            4

                                                               7        2                                     1

注意，如果从根到某个叶节点的路径上有的结点没有再输入中给出，或者给出了超过一次，应当输出-1.结点个数不超过256.

样例输入：

(11，LL)  (7，LLL)  (8，R)  (5，)  (4，L)  (13，RL)  (2，LLR)  (1，RRR)  (4，RR)  ()

(3，L)  (4，R) ()

样例输出：

5 4 8 11 13 4 7 2 1

-1

分析：

首先，看输入的结构解决方法 while(read_input(fp))  来解决，当读到EOF时返回0 来结束

然后，读取的方式（看程序）

最后就是注意里面的提示，在输出函数里面解决

下面是链式和数组两种方式实现程序：

链式：

#include<stdio.h>
#include<malloc.h>
#include<queue>
#include<string.h>

const int MAXN = 256;

typedef  struct TNode
{
	int hava_value;
	int v;
	struct TNode *left,*right;
} Node;

Node *root;

Node *newnode()
{
	Node * u = (Node *)malloc(sizeof(Node));
	if(u != NULL)
	{
		u->hava_value =  0;
		u->v = 0;
		u->left = u->right = NULL;
	}
	return u;
} 

void addnode(int v, char *s)
{
	int n = strlen(s);
	Node *u = root;
	for(int i = 0; i < n-1; ++i)
	{
		if(s[i] == 'L')
		{
			if(u->left == NULL)
				u->left = newnode();
			u = u->left;
		}
		else if(s[i]=='R')
		{
			if(u->right == NULL)
				u->right = newnode();
			u = u->right;
		}
	}
	if(u->hava_value)
	{
		printf("the value is existed\n");
		return;
	}
	u->v = v;
	u->hava_value = 1;
	
}


char s[MAXN + 10];

int read_input(FILE *fp)
{
	root = newnode();
	while(1)
	{
		if(fscanf(fp,"%s",s) != 1)
			return 0;
		if(!strcmp(s,"()"))
			break;
		int v;
		sscanf(&s[1],"%d",&v);
		addnode(v,strchr(s,',')+1);
	}
	return 1;
}

int bfs()
{
	using namespace std;
	int n = 0;
	int ans[MAXN];
	int fail = 0;
	
	queue<Node *> q;
	q.push(root);
	while(!q.empty())
	{
		Node * tmp = q.front();
		if(tmp->hava_value == 0)
		{
			fail = 1;
			break;
		}
		ans[n++] = tmp->v;
		if(tmp->left)
			q.push(tmp->left);
		if(tmp->right)
			q.push(tmp->right);
		q.pop();
	}
	if(fail == 1)
		printf("-1");
	else
		for(int i = 0; i < n; ++i)
			printf("%d ",ans[i]);
	printf("\n");
	return 1;
}

void remove_tree(Node *u)
{
	if(u == NULL)
		return;
	remove_tree(u->left);
	remove_tree(u->right);
	free(u);
}

int main()
{
	FILE *fp;
	fp = fopen("6.3.2.txt","r");
	if(!fp)
	{
		printf("File  opens  failly\n");
		return 1;
	}
	while(read_input(fp))
	{
		bfs();
		remove_tree(root);
	}
	fclose(fp);
	return 1;
}

数组：

#include<stdio.h>
#include<malloc.h>
#include<queue>
#include<string.h>

const int MAXN = 256;

typedef  struct TNode
{
	int hava_value;
	int v;
	int left,right;
} Node;

Node tree[MAXN];
int cnt = 1;

const int root = 1;

int newnode()
{
	int u = cnt++;
	tree[u].hava_value = 0;
	tree[u].v = 0;
	tree[u].left = tree[u].right = 0; 
	return u;
} 

void addnode(int v, char *s)
{
	int n = strlen(s);
	int u = root;
	for(int i = 0; i < n-1; ++i)
	{
		if(s[i] == 'L')
		{
			if(tree[u].left == 0)
				tree[u].left = newnode();
			u = tree[u].left;
		}
		else if(s[i] == 'R')
		{
			if(tree[u].right == 0)
				tree[u].right = newnode();
			u = tree[u].right;
		}
	}
	if(tree[u].hava_value)
	{
		printf("the value of %d is existed\n",v);
		return;
	}
	tree[u].v = v;
	tree[u].hava_value = 1;
	
}

void newtree()
{
	tree[root].left = tree[root].right = 0;
	tree[root].v = 0;
	tree[root].hava_value = 0;
	cnt = 2;
}

char s[MAXN + 10];
int read_input(FILE *fp)
{
	newtree();
	while(1)
	{
		if(fscanf(fp,"%s",s) != 1)
			return 0;
		if(!strcmp(s,"()"))
			break;
		int v;
		sscanf(&s[1],"%d",&v);
		addnode(v,strchr(s,',')+1);
	}
	return 1;
}

int bfs()
{
	using namespace std;
	int n = 0;
	int ans[MAXN];
	int fail = 0;
	
	queue<int> q;
	q.push(root);
	while(!q.empty())
	{
		int tmp = q.front();
		if(tree[tmp].hava_value == 0)
		{
			fail = 1;
			break;
		}
		ans[n++] = tree[tmp].v;
		if(tree[tmp].left != 0)
			q.push(tree[tmp].left);
		if(tree[tmp].right!= 0)
			q.push(tree[tmp].right);
		q.pop();
	}
	if(fail == 1)
		printf("-1");
	else
		for(int i = 0; i < n; ++i)
			printf("%d ",ans[i]);
	printf("\n");
	return 1;
}


int main()
{
	FILE *fp;
	fp = fopen("6.3.2.txt","r");
	if(!fp)
	{
		printf("File  opens  failly\n");
		return 1;
	}
	while(read_input(fp))
	{
		bfs();
	}
	fclose(fp);
	return 0;
}