1110 Complete Binary Tree (25 分)(搜尋)
阿新 • • 發佈:2018-12-01
1110 Complete Binary Tree (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 -
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
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程式碼:
#include<bits/stdc++.h> using namespace std; struct Node{ int data; int l,r; }gg[25]; int flag = 0, maxn = -1,ans; void DFS(int root, int id) { if(id > maxn){ maxn = id; ans = root; } if(gg[root].l != -1) DFS(gg[root].l, id * 2); if(gg[root].r != -1) DFS(gg[root].r, id * 2 + 1); return ; } int main() { int n; char s1[5],s2[5]; int b[22] = {0}; scanf("%d", &n); for(int i = 0 ;i < n; i++) { scanf("%s %s",s1,s2); if(s1[0] == '-') gg[i].l = -1; else { int data = 0, now = 0; while(s1[now] != '\0') data = data * 10 + s1[now++]-'0'; b[data] = 1; gg[i].l = data; } if(s2[0] == '-') gg[i].r = -1; else { int data = 0, now = 0; while(s2[now] != '\0') data = data * 10 + s2[now++]-'0'; b[data] = 1; gg[i].r = data; } gg[i].data = i; } int root = 0; for(int i = 0; i < n; i ++) { if(b[i] == 0) { root = i; break; } } DFS(root, 1); if(maxn == n) printf("YES %d\n",ans); else printf("NO %d\n",root); return 0; }