Given a tree, you are supposed to tell if it is a complete binary tree.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (≤20) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N−1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a - will be put at the position. Any pair of children are separated by a space.
Output Specification:
For each case, print in one line YES and the index of the last node if the tree is a complete binary tree, or NO and the index of the root if not. There must be exactly one space separating the word and the number.
Sample Input 1:
9
7 8
- -
- -
- -
0 1
2 3
4 5
- -
- -
Sample Output 1:
YES 8
Sample Input 2:
8
- -
4 5
0 6
- -
2 3
- 7
- -
- -
Sample Output 2:
NO 1
1 #include <stdio.h> 2 #include <algorithm> 3 #include <set> 4 #include <string.h> 5 #include <vector> 6 #include <math.h> 7 #include <queue> 8 #include <iostream> 9 #include <string> 10 using namespace std; 11 const int maxn = 30; 12 int n,k; 13 struct node{ 14 int l=-1,r=-1; 15 }tree[maxn]; 16 int vis[maxn]={0}; 17 int main(){ 18 scanf("%d",&n); 19 getchar(); 20 for(int i=0;i<n;i++){ 21 string c1,c2; 22 cin>>c1>>c2; 23 getchar(); 24 int n1=-1,n2=-1; 25 if(c1!="-"){ 26 n1=stoi(c1); 27 vis[n1]=1; 28 } 29 if(c2!="-"){ 30 n2=stoi(c2); 31 vis[n2]=1; 32 } 33 tree[i].l=n1; 34 tree[i].r=n2; 35 } 36 int root=-1; 37 for(int i=0;i<n;i++){ 38 if(vis[i]==0){ 39 root=i; 40 break; 41 } 42 } 43 int flag=0,flag2=0,now; 44 queue<int> q; 45 q.push(root); 46 while(!q.empty()){ 47 now=q.front(); 48 q.pop(); 49 //printf("%d ",now); 50 if(tree[now].l!=-1){ 51 if(flag==0)q.push(tree[now].l); 52 else { 53 flag2=1; 54 break; 55 } 56 } 57 else{ 58 flag=1; 59 } 60 if(tree[now].r!=-1){ 61 if(flag==0)q.push(tree[now].r); 62 else { 63 flag2=1; 64 break; 65 } 66 } 67 else{ 68 flag=1; 69 } 70 /*if(now!=-1){ 71 flag++; 72 flag2=now; 73 } 74 else{ 75 if(flag==n){ 76 printf("YES %d",flag2); 77 } 78 else{ 79 printf("NO %d",root); 80 } 81 return 0; 82 } 83 q.push(tree[now].l); 84 q.push(tree[now].r); 85 */ 86 } 87 if(flag2==1)printf("NO %d",root); 88 else printf("YES %d",now); 89 }
注意点：完全二叉树的判断就是看已经有节点没孩子了，后面节点却还有孩子，这树就不是完全二叉树。有三个点一直错误，一开始段错误，后来在结构体初始化了-1后答案错误，总找不到原因。后来才发现原来是读输入的时候错了，两位数就读不到了，不能用%c，要用string