Trees on the level

最新推荐文章于 2021-07-25 01:08:48 发布

原创最新推荐文章于 2021-07-25 01:08:48 发布 · 215 阅读

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19暑假集训同时被 3 个专栏收录

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本文探讨了二叉树的层次遍历算法，通过解析输入的二叉树序列，实现树的构建与遍历，并处理不完全指定的树结构。文章提供了完整的代码示例，展示了如何使用C++进行层次遍历的实现。

题目描述：

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 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
在这里插入图片描述
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.

输入：

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.

输出：

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

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
not complete

作为一只蒟蒻的我，看到判断是否是树，总是莫名其妙想到的并查集？？？
但是仔细阅读，发现，，虽然也不是不能写，就是，，太麻烦啦，估计要好长好长
巨懒的我觉定看一下别人的思路，顺便学习一下
下面是总结之后的代码

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<vector>
#include<map> 
using namespace std;
struct node{
	int x,y;
	string s;
}edge[1010];
map<string,int> feng;
bool cmp(node a,node b)
{
	if(a.y==b.y)return a.s<b.s;
	else return a.y<b.y;
}
int main()
{
	string a;
	int n=0,flag=0;
	while(cin>>a)
	{
		if(a=="()")
		{
			sort(edge,edge+n,cmp);
			if(flag==1||edge[0].y!=0)
			{
				printf("not complete\n");
			}
			else
			{
				for(int i=0;i<n;i++)
				{
					if(edge[i].y>0)
					{
						string ss="";
						int len=edge[i].s.length();
						for(int j=0;j<len-1;j++)
						{
							ss+=edge[i].s[j];
						}
						if(feng[ss]!=1)
						{
							flag=1;
							break;
						}
					}
				}
				if(flag==1)printf("not complete\n");
				else
				{
					for(int i=0;i<n;i++)
					{
						if(i!=0)printf(" ");
						printf("%d",edge[i].x);
					}
					printf("\n");
				 } 
			}
			n=flag=0;
			feng.clear();
		}
		else
		{
			int i;
			int num=0;
			for(i=1;i<a.length();i++)
			{
				if(a[i]==',')break;
				else
				{
					num=num*10+a[i]-'0';
				}
			}
			edge[n].x=num;
			edge[n].s.clear();
			num=0;
			for(i=i+1;i<a.length();i++)
			{
				if(a[i]==')')break;
				num++;
				edge[n].s+=a[i];
			}
			feng[edge[n].s]++;
			if(feng[edge[n].s]>1)flag=1;
			edge[n].y=num;
			n++;
		}
	} 
	return 0;
}