2020牛客寒假算法基础集训营3 D.牛牛与二叉树的数组存储

D.牛牛与二叉树的数组存储

题目链接-D.牛牛与二叉树的数组存储


上图是一颗满二叉树，对于每个节点i，i2是它的左子树，i2+1是它的右子树，i/2则是它的父亲节点。当然了，当树不为满二叉树的时候其实也能储存
上图是树不为满二叉树的情况，我们先用一些虚点将其补充成一颗满二叉树。
根节点还是被储存到数组的第1个位置。
然后对于下标为i的节点，他的左孩子的下标为i* 2，它的右孩子的下标为i*2+1，它的父亲节点的下标为i/2。
给你一个长度为n的数组，该数组储存了一颗二叉树，数组中仅含有-1和正整数，且整个数组中的正整数是从1到树尺寸连续的，不会出现如1,2,3,5,6，这种缺少4断掉的情况。
请你告诉我树的尺寸和根节点，并按照顺序打印每个节点的父亲节点、左孩子、右孩子分别是谁？
输入

7
1 2 3 4 5 6 7

输出

The size of the tree is 7
Node 1 is the root node of the tree
The father of node 1 is -1, the left child is 2, and the right child is 3
The father of node 2 is 1, the left child is 4, and the right child is 5
The father of node 3 is 1, the left child is 6, and the right child is 7
The father of node 4 is 2, the left child is -1, and the right child is -1
The father of node 5 is 2, the left child is -1, and the right child is -1
The father of node 6 is 3, the left child is -1, and the right child is -1
The father of node 7 is 3, the left child is -1, and the right child is -1

输入

7
3 -1 2 -1 -1 1 4

输出

The size of the tree is 4
Node 3 is the root node of the tree
The father of node 1 is 2, the left child is -1, and the right child is -1
The father of node 2 is 3, the left child is 1, and the right child is 4
The father of node 3 is -1, the left child is -1, and the right child is 2
The father of node 4 is 2, the left child is -1, and the right child is -1

解题思路
水题，按题目要求模拟即可

附上代码

#include<bits/stdc++.h>
using namespace std;
const int N=4e6+5;
const int M=1e9+7;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
typedef long long ll;
typedef pair<int,int> PII;
int a[N],pos[N];
int main(){
	
	int n;
	cin>>n;
	memset(a,-1,sizeof a);
	int cnt=0;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		if(a[i]!=-1) 
			pos[a[i]]=i,cnt++;
	}
	printf("The size of the tree is %d\n",cnt);
	printf("Node %d is the root node of the tree\n",a[1]);
	for(int i=1;i<=cnt;i++){
		printf("The father of node %d is %d, the left child is %d, and the right child is %d\n",i,a[pos[i]/2],a[pos[i]*2],a[pos[i]*2+1]);
	}
	return 0;
}