题解
- 给定序列,我们建立一颗二叉搜索树,这里用数组存树(根节点为n,那么左儿子为2n,右儿子为2n+1),建树过程,我们每次从根节点开始遍历,遇到大于根节点的就向左递归,否则向右递归,同时我们更新maxn(用来记录最后一个节点所在数组的下标),这样一颗树就建好了
- 对于层序遍历,我们直接按数组下标输出即可,判断是否为完全二叉树,我们只需要看前n个点是否在数组前n个位置即可
代码
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1e5 + 10;
int n, x;
int tree[N];
int main() {
cin >> n;
int maxn = 0;
for (int i = 0; i < n; i++) {
cin >> x;
int now = 1;
while (tree[now]) {
if (x > tree[now]) {
now = now * 2;
} else {
now = now * 2 + 1;
}
}
tree[now] = x;
if (now > maxn) maxn = now;
}
bool flag = true;
cout << tree[1];
for (int i = 2; i <= maxn; i++) {
if (tree[i]) cout << " " << tree[i];
else flag = false;
}
cout << endl;
if (flag) cout << "YES" << endl;
else cout << "NO" << endl;
return 0;
}