树的层次遍历（Trees on the level，UVA 122）

题目描述：

题目思路：

1.用结构链表来建树

2.用队列来实现层次遍历，当遍历到根节点时，将其子节点压入队列

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
 
//树结点
struct Node{
    int v ;
    Node* left,*right ;
    int have_value ;
    Node():have_value(false),left(NULL),right(NULL){} ;
} ;
 
Node* root ;//根节点
 
Node* newnode(){
    return new Node() ; //返回一个新结点
} 
 
bool failed ;
 
void addnode(int v,char* s){//添加新结点
    int n = strlen(s);
    Node* u = root ;
     for(int i = ;i < n;i++)//找到要加入的位置
     {
         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 ;//是否已经被访问过；
     u->v = v;
     u->have_value = true;
} 
 
void freetree(Node* u){ //释放内存
    if(u == NULL) return ;
    freetree(u->left);
    freetree(u->right);
    delete u;
}
 
char s[];
bool read_input(){
    failed = false ;
    freetree(root) ;
    root = newnode();
    while(true){
        if(scanf("%s", s) != ) return false;
        if(!strcmp(s,"()")) break;
        int v ;
        sscanf(&s[],"%d",&v);
        addnode(v,strchr(s,',')+);
    }
    return true ;
} 
 
bool bfs(vector<int>& ans){//搜索
    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->v);
        if(u->left != NULL)    q.push(u->left);
        if(u->right != NULL) q.push(u->right);
    }
    return true ;
}
 
int main(int argc, char *argv[])
{
    vector<int> ans;
    while(read_input()){
        if(!bfs(ans)) failed = ;
        if(failed) printf("not complete\n");
        else{
            for(int i = ;i < ans.size();i++)
            {
                if(i != ) cout << " " ;
                   cout << ans[i];
            }
            cout << endl ;
        }
    }
    return ;
}