Trees on the Level

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Trees are fundamental in many branches of computer science. Current state-of-the art parallel computers such as Thinking Machines' CM-5 are based on fat trees. Quad- and octal-trees are fundamental to many algorithms in computer graphics.

This problem involves building and traversing binary trees.

Given a sequence of binary trees, you are to write a program that prints a level-order traversal of each tree. In this problem each node of a binary tree contains a positive integer and all binary trees have fewer than 256 nodes.

In a level-order traversal of a tree, the data in all nodes at a given level are printed in left-to-right order and all nodes at level k are printed before all nodes at level k+1.

For example, a level order traversal of the tree

is: 5, 4, 8, 11, 13, 4, 7, 2, 1.

In this problem a binary tree is specified by a sequence of pairs (n,s) where n is the value at the node whose path from the root is given by the string s. A path is given be a sequence of L's and R's where L indicates a left branch and R indicates a right branch. In the tree diagrammed above, the node containing 13 is specified by (13,RL), and the node containing 2 is specified by (2,LLR). The root node is specified by (5,) where the empty string indicates the path from the root to itself. A binary tree is considered to be completely specified if every node on all root-to-node paths in the tree is given a value exactly once.

Input

The input is a sequence of binary trees specified as described above. Each tree in a sequence consists of several pairs (n,s) as described above separated by whitespace. The last entry in each tree is (). No whitespace appears between left and right parentheses.

All nodes contain a positive integer. Every tree in the input will consist of at least one node and no more than 256 nodes. Input is terminated by end-of-file.
Output

For each completely specified binary tree in the input file, the level order traversal of that tree should be printed. If a tree is not completely specified, i.e., some node in the tree is NOT given a value or a node is given a value more than once, then the string ``not complete'' should be printed.

Sample Input

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

Sample Output

5 4 8 11 13 4 7 2 1
not complete

【分析】题意就是，根据输入，输出树的层序遍历或者 not complete。输入以括号结束。

【代码】参考了书上的题解和代码....

#include<bits/stdc++.h>
using namespace std;
const int maxn=256+10;
char s[maxn];
struct node{
	node *left;
	node *right;
	int val;
	bool have_value;
}*root;
bool failed;//插入是否成功
node *newnode()//新节点的建立
{
	return new node(); 
}
void addnode(int val,char s[])
{
	int len=strlen(s);
	node *u=root;////
	for(int i=0;i<len;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->have_value)failed=true;//如果已经赋过值了，标记插入失败
	u->val=val;
	u->have_value=true;
}
bool readin()
{
	failed=false;
	while(1)
	{
		if(scanf("%s",s)!=1)return false;
		if(!strcmp(s,"()"))break;
		int x;
		sscanf(&s[1],"%d",&x);//字符串转换为整数。可以把任意指向字符的指针看成是字符串，从该位置开始一直到'\0'；
		addnode(x,strchr(s,',')+1);
	}
	return true;
}
bool bfs(vector<int>&ans)
{
	queue<node*>q;
	ans.clear();
	q.push(root);
	while(!q.empty())
	{
		node *u=q.front();
		q.pop();
		if(!u->have_value)return false;
		ans.push_back(u->val);
		if(u->left!=NULL)q.push(u->left);
		if(u->right!=NULL)q.push(u->right);
	}
	return true;
}
int main()
{
	while(1)
	{
		root=newnode();
		if(!readin())break;
		vector<int>ans;
		if(!failed&&bfs(ans))
		{
			for(int i=0;i<ans.size();i++)
			printf("%d%c",ans[i],i==ans.size()-1?'\n':' ');
		}
		else puts("not complete");
	}
	return 0;
}