题目链接：
https://cn.vjudge.net/problem/UVA-122
1 /* 2 问题 3 给出每个节点的权值和路线，输出该二叉树的层次遍历序列。 4 5 解题思路 6 根据输入构建链式二叉树，再用广度优先遍历保存权值最后输出。 7 */ 8 #include<cstdio> 9 #include<cstring> 10 #include<vector> 11 #include<queue> 12 using namespace std; 13 const int maxn=110; 14 bool failed; 15 16 struct NODE{ 17 bool have_value; 18 int v; 19 NODE *left,*right; 20 NODE() : have_value(false),left(NULL),right(NULL){}; 21 }; 22 NODE* newnode(){ 23 return new NODE(); 24 } 25 NODE* root; 26 27 bool read_input(); 28 void addnode(int v,char *s); 29 bool bfs(vector<int> &ans); 30 void remove_tree(NODE* u){ 31 if(u == NULL) return; 32 remove_tree(u->left); 33 remove_tree(u->right); 34 delete u; 35 } 36 37 int main() 38 { 39 //freopen("E:\testin.txt","r",stdin); 40 vector<int> ans; 41 while(read_input()){ 42 if(failed || !bfs(ans)) 43 printf("not complete "); 44 else{ 45 int i; 46 for(i=0;i<ans.size()-1;i++) 47 printf("%d ",ans[i]); 48 printf("%d ",ans[i]); 49 } 50 } 51 return 0; 52 } 53 54 bool bfs(vector<int> &ans){ 55 queue<NODE*> q; 56 ans.clear(); 57 q.push(root); 58 59 while(!q.empty()){ 60 NODE* u =q.front(); q.pop(); 61 if(!u->have_value) return false; 62 ans.push_back(u->v); 63 64 if(u->left != NULL) q.push(u->left); 65 if(u->right != NULL) q.push(u->right); 66 } 67 return true; 68 } 69 void addnode(int v,char *s){ 70 int len=strlen(s); 71 72 NODE* u= root; 73 for(int i=0;i<len;i++){ 74 if(s[i] == 'L'){ 75 if(u->left == NULL) 76 u->left= newnode(); 77 u=u->left; 78 }else if(s[i] == 'R'){ 79 if(u->right == NULL) 80 u->right= newnode(); 81 u=u->right; 82 } 83 } 84 85 if(u->have_value) failed=true; 86 u->v= v; 87 u->have_value=true; 88 } 89 bool read_input(){ 90 char s[maxn]; 91 failed=false; 92 remove_tree(root); 93 root=newnode(); 94 for(;;){ 95 if(scanf("%s",s) != 1) return false; 96 if(!strcmp(s,"()")) break; 97 int v; 98 sscanf(s+1,"%d",&v); 99 addnode(v,strchr(s,',')+1); 100 } 101 return true; 102 }