PAT 1110 Complete Binary Tree[比较]
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 -
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
题目大意:给出一个二叉树,判断是否是完全二叉树。
//这个题真的学到了不少东西。我的AC:
参照了:https://blog.csdn.net/hyf20144055065/article/details/51970789
#include <iostream>
#include <algorithm>
#include<cstdio>
#include<stdio.h>
#include <queue>
#include<cmath>
#include <vector>
using namespace std;
struct Node{
int left=-,right=-;
}node[];
int isr[];
int root=-;
int ct=,n,last=;
void level(int r){//进行层次遍历。我是选择在入队的时候+,但是这样是比较复杂的,我应该在出队的时候+。
queue<int> qu;
qu.push(r);
//ct++;
while(!qu.empty()){
//if(ct==n)last=qu.back();
int top=qu.front();qu.pop();
if(top==-)break;
last=top;//每次last都赋值为当前。
ct++;
// if(node[top].left==-1)break;
// else {
// qu.push(node[top].left);
// ct++;
// }
qu.push(node[top].left);//其实这里-1完全可以push进去,因为下一次再次弹出时会进行判断的。
qu.push(node[top].right);
// if(node[top].left!=-1){
// qu.push(node[top].right);
// ct++;
// }
}
} int main()
{
cin>>n;
string l,r;
for(int i=;i<n;i++){
cin>>l>>r;
if(l!="-"){
node[i].left=atoi(l.c_str());//注意这里的转换
//cout<<node[i].left<<'\n';
isr[node[i].left]=;
}
if(r!="-"){
node[i].right=atoi(r.c_str());
isr[node[i].right]=;
}
}
for(int i=;i<n;i++){
if(isr[i]==){
root=i;break;
}
}
level(root);
if(ct==n)
cout<<"YES "<<last;
else
cout<<"NO "<<root;
return ;
}
1.利用完全二叉树的性质来判断。
2。使用了层次遍历,但是我的层次遍历思路是不正确的。
3.我的思路:只要有左子节点就push进去,然后计数+1,是在入队的时候计算数量。只要左子节点为-1,那么就break掉。
4.正确思路:当节点左右为空-1时,可以push进去,在whlle循环中会进行判断,如果是-1,那么就break了,是在出队的时候进行+1操作,这样最后就可以了
5.在我原来的那种方法中,2和6测试点过不去,对于6测试点:
使用了
1
- -
测试数据进行了更正
2测试点是对于0的检测。
正确输出应该是YES 0
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