PAT (Advanced Level) Practise 1110 Complete Binary Tree (25)

1110. Complete Binary Tree (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

97 8- -- -- -0 12 34 5- -- -

Sample Output 1:

YES 8

Sample Input 2:

8- -4 50 6- -2 3- 7- -- -

Sample Output 2:

NO 1

题意：一棵树有n个节点，告诉你各个节点的左右孩子，问这棵树是不是完全二叉树，若是，输出最后一个点的编号，若不是，输出根节点

解题思路：首先建树后，没有作为叶子节点的一定是根节点，然后算所有点的标号，若最大为n，则是完全二叉树

#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <queue>#include <vector>#include <set>#include <stack>#include <map>#include <climits>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;int n, son[25][2],visit[25];int main(){while (~scanf("%d", &n)){memset(visit, 0, sizeof visit);for (int i = 0; i < n; i++){char s1[5], s2[5];scanf("%s%s", &s1, &s2);int a = 0, b = 0,len1=strlen(s1),len2=strlen(s2);if (s1[0] != '-'){for (int j = 0; j < len1; j++) a = a * 10 + s1[j] - '0';son[i][0] = a;visit[a] = 1;}else son[i][0] = -1;if (s2[0] != '-'){for (int j = 0; j < len2; j++) b = b * 10 + s2[j] - '0';son[i][1] = b;visit[b] = 1;}else son[i][1] = -1;}int ma = 1,k,ans;for (int i = 0; i < n; i++)if (!visit[i]) k = i;queue<pair<int, int> >q;q.push(make_pair(k, 1));while (!q.empty()){pair<int, int>pre = q.front();                        q.pop();if (pre.second == n) ans = pre.first;ma = max(ma, pre.second);if (son[pre.first][0] != -1) q.push(make_pair(son[pre.first][0],pre.second*2));if (son[pre.first][1] != -1) q.push(make_pair(son[pre.first][1], pre.second * 2+1));}if (ma == n) printf("YES %d\n", ans);else printf("NO %d\n", k);}return 0;}