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:

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 long
const 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;
}