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 8Sample Input 2:
8 - - 4 5 0 6 - - 2 3 - 7 - - - -Sample Output 2:
NO 1
判断是否是完全二叉树,用字符串读入结点,然后转换,层序遍历,遇到儿子是空的就停止,看看队列里是否是n个结点。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int n,v[30],root = -1,q[30],head = 0,tail = 0; char l[3],r[3]; struct Node { int left,right; }s[30]; int main() { cin>>n; for(int i = 0;i < n;i ++) { cin.get(); cin>>l; if(strcmp(l,"-") == 0)s[i].left = -1; else { int d = 0; for(int j = 0;l[j];j ++) d = d * 10 + l[j] - '0'; s[i].left = d; v[d] = 1; } cin.get(); cin>>r; if(strcmp(r,"-") == 0)s[i].right = -1; else { int d = 0; for(int j = 0;r[j];j ++) d = d * 10 + r[j] - '0'; s[i].right = d; v[d] = 1; } } for(int i = 0;i < n;i ++) if(!v[i]) { root = i; break; } q[tail ++] = root; while(head < tail) { if(s[q[head]].left != -1)q[tail ++] = s[q[head]].left; else break; if(s[q[head]].right != -1)q[tail ++] = s[q[head]].right; else break; head ++; } if(tail == n)cout<<"YES "<<q[tail - 1]; else cout<<"NO "<<root; }