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
思路:
给出一个n表示有n个结点,这n个结点为0~n-1,给出这n个结点的左右孩子,求问这棵树是不是完全二叉树
C++:
#include "cstdio"
#include "iostream"
#include "queue"
#include "string"
using namespace std;
struct node
{
int lchild,rchild;
};
node tree[25];
int book[25]={0};
int main(){
int n;
string temp1,temp2;
scanf("%d",&n);
for (int i=0;i<n;i++)
{
cin>>temp1>>temp2;
getchar();
if (temp1=="-")
{
tree[i].lchild=-1;
}else
{
tree[i].lchild=stoi(temp1);
book[stoi(temp1)]=1;
}
if (temp2=="-")
{
tree[i].rchild=-1;
}else
{
tree[i].rchild=stoi(temp2);
book[stoi(temp2)]=1;
}
}
//记录根结点
int root=-1;
for (int i=0;i<n;i++)
{
if(book[i]==0){root=i;break;}
}
queue<int> q;
q.push(root);
int cnt = 0, lastnode = 0;
while(!q.empty()) {
int temp = q.front();
q.pop();
if(temp != -1) {
lastnode = temp;
cnt++;
}else {
if(cnt != n)
printf("NO %d", root);
else
printf("YES %d", lastnode);
return 0;
}
q.push(tree[temp].lchild);
q.push(tree[temp].rchild);
}
return 0;
return 0;
}