UVa 122 - Trees on the level【二叉树初接触（BFS）】

122 - Trees on the level

Background

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.

The Problem

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

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 Lindicates 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 becompletely specified if every node on all root-to-node paths in the tree is given a value exactly once.

The 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.

The 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 1not complete

#include<iostream>#include<vector>#include<cstdio>#include<new>#include<cstring>#include<queue>#define maxn 100000using namespace std;struct Node{    int v;//节点值    bool have_value;//是否被赋值过    Node *left,*right;    Node():have_value(false),left(NULL),right(NULL){};};Node* root;char s[maxn];//保存输入的节点bool failed=false;void addnode(int v,char *s){    int n=strlen(s);    Node* u=root;//从根节点开始往下走    for(int i=0;i<n;i++){        if(s[i]=='L'){            if(u->left==NULL)u->left=new Node();//节点不存在，创建新节点            u=u->left;        }        else if(s[i]=='R'){            if(u->right==NULL)u->right=new Node();//节点不存在，创建新节点            u=u->right;        }//忽略其他情况，即最后那个多余的又括号    }    if(u->have_value) failed=true;    u->v=v;    u->have_value=true;}bool read_input(){    failed=false;    root=new Node();//创建根节点    for(;;){        if(scanf("%s",s)!=1) return false;//输入结束        if(!strcmp(s,"()"))break;//读到结束标志，退出循环        int v;        sscanf(&s[1],"%d",&v);//读入节点值        addnode(v,strchr(s,',')+1);//查找逗号，然后插入节点（函数strchr（s，‘，’）返回字符串s中从左往右第一个‘，’的指针）    }    return true;}bool bfs(vector<int>& ans){    queue<Node*> q;    ans.clear();    q.push(root);//初始时只有一个节点    while(!q.empty()){        Node *u=q.front();        q.pop();        if(!u->have_value) return false;//有节点没有被赋值过，表明输入错误        ans.push_back(u->v);//增加到输出序列尾部        if(u->left!=NULL) q.push(u->left);//把左子节点（如果有）放入队列；        if(u->right!=NULL) q.push(u->right);//把右子节点（如果有）放入队列；    }    return true;//输入正确；}int main(){    while(read_input()){        if(failed)            printf("not complete\n");        else{            vector<int>ans;            if(bfs(ans)){                for(vector<int>::iterator i=ans.begin();i!=ans.end();i++){                    if(i!=ans.begin())                        printf(" ");                    printf("%d",*i);                }                printf("\n");            }            else printf("not complete\n");        }    }    return 0;}