PAT A1110 Complete Binary Tree (25point(s))

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 1：类似于：已知树形（完全二叉树：id ：左 2*id, 右 2*id+1）+ 一个遍历序列 =》 得到树 的问题
  静态存储树后先序遍历，将节点存入一个数组（将下标作为参数传入PreOrder() )
  遍历数组判断是否前n个元素都装满

• TIPS：节点编号从0开始连续：求根
  可以先求出0 + 1 + 2 + … + n-1 = (n - 1) * n / 2 并减去空指针（-1）
  然后每得到一个孩子节点的编号就减去这个编号（因为空指针为-1，已经处理过，不用单独考虑）

• code :

#include <bits/stdc++.h>
using namespace std;
const int maxn = 50;
bool not_root[maxn];
struct Node{
	int lc, rc;
}node[maxn];
int GetRoot(int n){
	for(int i = 0; i < n; ++i) if(not_root[i] == false) return i;
}
int tree[maxn];
void PreOrder(int r, int id){
	if(r == -1) return;
	tree[id] = r;
	PreOrder(node[r].lc, 2 * id);
	PreOrder(node[r].rc, 2 * id + 1);
}
int main(){
	int n;
	scanf("%d", &n);
	int r = (n - 1) * n / 2 - (n + 1);	//所有节点编号和-空指针数 
	for(int i = 0; i < n; ++i){
		getchar();
		string tl, tr;
		cin >> tl >> tr;
		node[i].lc = tl[0] == '-' ? -1 : stoi(tl);
		node[i].rc = tr[0] == '-' ? -1 : stoi(tr);		
		r -= (node[i].lc + node[i].rc);
	}	
	memset(tree, -1, sizeof(tree));
	PreOrder(r, 1);
	for(int i = 1; i <= n; ++i){
		if(tree[i] == -1){
			printf("NO %d", r);
			return 0;
		}
	}
	printf("YES %d", tree[n]);
	return 0;
}