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 else
33 {
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 else
78 {
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。