uva 122 trees on the level——yhx

题目如下：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 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.

 1 #include<cstdio> 2 #include<cstring> 3 #include<queue> 4 using namespace std;  5 struct node 6 { 7     int lch,rch,val; 8     bool b; 9 }a[260],n1,n2;10 int n,ans[260];11 int rd()12 {13     int i,j,k,p,x,y,z,rt;14     char s[300],c1,c2;15     memset(a,0,sizeof(a));16     n=1;17     if (scanf("%s",s)==-1) return 0;18     rt=1;19     while (1)20     {21         if (s[1]==')') break;22         x=0;23         for (i=1;s[i]!=',';i++)24           x=x*10+s[i]-'0';25         p=1;26         for (i=i+1;i<=strlen(s)-2;i++)27           if (s[i]=='L')28           {29               if (!a[p].lch) a[p].lch=++n;30               p=a[p].lch;31           }32           else33           {34               if (!a[p].rch) a[p].rch=++n;35               p=a[p].rch;36           }37         if (a[p].b) rt=-1;38         a[p].val=x;39         a[p].b=1;40         scanf("%s",s);41     }42     return rt;43 }44 queue<int> q;45 int main()46 {47     int i,j,k,l,m,p,x,y,z;48     bool b;49     while (1)50     {51         x=rd();52         if (!x) break;53         if (x==-1) 54         {55             printf("not complete\n");56             continue;57         }58         while (!q.empty()) q.pop();59         q.push(1);60         memset(ans,0,sizeof(ans));61         k=b=0;62         while (!q.empty())63         {64             n1=a[q.front()];65             q.pop();66             if (!n1.b)67             {68                 b=1;69                 break;70             }71             ans[++k]=n1.val;72             if (n1.lch) q.push(n1.lch);73             if (n1.rch) q.push(n1.rch);74         }75         if (b)76           printf("not complete\n");77         else78         {79             printf("%d",ans[1]);80             for (i=2;i<=k;i++)81                  printf(" %d",ans[i]);82                printf("\n");83         }84     }85 }

如果直接用数组下标表示位置，那么当所有节点连成一条线的时候下标将变得很大。所以需要在每个节点记录他的左右儿子位置，这样可以节省空出来的空间。

每个节点用一个bool记录是否被赋过值，如果没被赋过值或是被赋第二次值，那就not complete了。

但是注意的细节就是由于多组数据，即使读入时已经知道not complete也要读完。