1110. Complete Binary Tree

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

先建树，把位置记下来，根节点的位置为0。

然后遍历一下，看看位置有没有超过n，顺便记一下位置为n-1的值。

#include<iostream>#include<algorithm>#include<vector>#include<string>using namespace std;typedef struct node{int pos;int l,r;};node t[20];bool f[20];void buildtree(int root,int pos){t[root].pos=pos;if(t[root].l!=-1)buildtree(t[root].l,2*pos+1);if(t[root].r!=-1)buildtree(t[root].r,2*pos+2);}int s2i(string s){if(s=="-")return -1;int num=0;for(int i=0;i<s.size();i++)num=num*10+s[i]-'0';f[num]=true;return num;}int main(){int n;cin>>n;for(int i=0;i<n;i++)f[i]=false;for(int i=0;i<n;i++){string l,r;cin>>l>>r;t[i].l=s2i(l);t[i].r=s2i(r);}int root;for(int i=0;i<n;i++)if(f[i]==false)root=i;buildtree(root,0);int rear;for(int i=0;i<n;i++){if(!(t[i].pos>=0&&t[i].pos<n)){cout<<"NO"<<" "<<root;return 0;}else if(t[i].pos==n-1)rear=i;}cout<<"YES"<<" "<<rear;return 0;}