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

问一棵树是不是完全二叉树，直接给根节点赋值1，然后左边是2，右边是3，这样判断最大的值是不是大于n即可

#include<cstdio>#include<vector>#include<queue>#include<string>#include<map>#include<iostream>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int INF = 0x7FFFFFFF;const int maxn = 1e5 + 10;int n, ch[maxn][2], flag[maxn], ans;char s[maxn];int get(){scanf("%s", s);return s[0] == '-' ? -1 : s[1] ? s[1] - '0' + 10 * (s[0] - '0') : s[0] - '0';}bool dfs(int x, int y){if (x == -1) return true;if (y > n) return false;if (y == n) ans = x;return dfs(ch[x][0], y << 1) && dfs(ch[x][1], y << 1 | 1);}int main(){scanf("%d", &n);for (int i = 0; i < n; i++){ch[i][0] = get();if (ch[i][0] != -1) flag[ch[i][0]] = 1;ch[i][1] = get();if (ch[i][1] != -1) flag[ch[i][1]] = 1;}for (int i = 0; i < n; i++){if (!flag[i]){if (dfs(i, 1)) printf("YES %d\n", ans);else printf("NO %d\n", i);}}return 0;}