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
在节点中为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 cin>>n; 9 vector<node> v(n); 10 vector<int> vis(n, 0); 11 string a, b; 12 for(i=0; i<n; i++){ 13 cin>>a>>b; 14 if(a=="-") v[i].left = -1; 15 else{ 16 v[i].left = stoi(a); 17 vis[v[i].left] = 1; 18 } 19 if(b=="-") v[i].right = -1; 20 else{ 21 v[i].right = stoi(b); 22 vis[v[i].right] = 1; 23 } 24 } 25 for(i=0; i<n; i++) 26 if(vis[i]==0) root=i; 27 queue<int> q; 28 q.push(root); 29 int temp, flag=0, f=0 ; 30 while(q.size()){ 31 temp = q.front(); 32 q.pop(); 33 if(f) break; 34 if(v[temp].left != -1){ 35 q.push(v[temp].left); 36 if(flag) f=1; 37 }else flag=1; 38 if(v[temp].right != -1){ 39 q.push(v[temp].right); 40 if(flag) f=1; 41 }else flag=1; 42 } 43 if(f) printf("NO %d", root); 44 else printf("YES %d", temp); 45 return 0; 46 }