1110 Complete Binary Tree (25)（25 分）
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
题目大意：给出一颗二叉树的结构，判断该二叉树是否是完全二叉树；
思路：用数组保存树的结构
在节点中未出现的值就是根节点
层序遍历二叉树， 如果已经出现一个空节点， 如果还出现非空节点， 则该树一定不是完全二叉树；
注意点：开始看输入样例， 误以为输入都是一个字符串，导致三个测试点不能通过， 找了很久的原因， 最多有20个节点， 用string输入就行
1 #include<iostream> 2 #include<vector> 3 #include<queue> 4 using namespace std; 5 struct node{ int left, right;}; 6 int main(){ 7 int n, root, i; 8 scanf("%d", &n); 9 vector<node> v(n); 10 //记录出现过的节点 11 vector<int> vis(n, 0); 12 string a, b; 13 for(i=0; i<n; i++){ 14 cin>>a>>b; 15 if(a=="-") v[i].left = -1; 16 else{ 17 v[i].left = stoi(a); 18 vis[v[i].left] = 1; 19 } 20 if(b=="-") v[i].right = -1; 21 else{ 22 v[i].right = stoi(b); 23 vis[v[i].right] = 1; 24 } 25 } 26 for(i=0; i<n; i++) //没有出现过在子节点中的点就是根节点 27 if(vis[i]==0) root=i; 28 queue<int> q; 29 q.push(root); 30 int temp, flag=0, f=0 ; 31 //flag:记录是否出现空节点 32 //f：记录出现空节点后， 是否还有后继的节点 33 while(q.size()){ 34 temp = q.front(); 35 q.pop(); 36 if(f) break; 37 if(v[temp].left != -1){ 38 q.push(v[temp].left); 39 if(flag) f=1; 40 }else flag=1; 41 if(v[temp].right != -1){ 42 q.push(v[temp].right); 43 if(flag) f=1; 44 }else flag=1; 45 } 46 if(f) printf("NO %d", root); 47 else printf("YES %d", temp); 48 return 0; 49 }