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.

输入：

9
7 8
- -
- -
- -
0 1
2 3
4 5
- -
- -

输出：

YES 8

输入：

8
- -
4 5
0 6
- -
2 3
- 7
- -
- -

输出：

NO 1

题意为给出一颗二叉树，第一行输入一个数n（数的结点数0~n-1),后面的n行分别是对应结点的左右子节点，-为空结点。然后要你判断这个数是不是完全二叉树，是的话输出YES和最后一个结点的下标，不是的话输出NO和根结点下标。

直接上代码，应该很容易看懂

#include<bits/stdc++.h>
using namespace std;
const int N = 100;
struct node {
	int index;
	int lchild, rchild;
	int nums;
	node() {
		lchild = rchild = -1;
	}
	node(int a) : nums(a) { }
}tree[N];
int n;
int max_nums = 0;
int get_id(string str) {
	int sum = 0, len = str.length();
	if(str[0] == '-') return -1;
	
	for(int i = 0; i < len; i++) {
		sum = sum*10 + str[i] - '0';
	}
	return sum;
} 
int find_farther[N];
void bfs(int s) {
	queue<node> q;
	q.push(tree[s]);
	int nums = 1;
	while(!q.empty()) { //给结点编号
		node now = q.front();
		q.pop();
		tree[now.index].nums = nums++;
		if(nums > n) max_nums = now.index; 
		if(now.lchild != -1) {
			q.push(tree[now.lchild]);
		} 
		 if(now.rchild != -1) {
			q.push(tree[now.rchild]);
		}
	}
}

bool dfs(int s) {  //判断是不是完全二叉树，就是左右孩子的编号是否合法
	if(s == -1) return true;
	if(tree[s].lchild != -1 && tree[s].nums*2 != tree[tree[s].lchild].nums) {
		return false;
	}
	if(tree[s].rchild != -1 && tree[s].nums*2+1 != tree[tree[s].rchild].nums) {
		return false;
	}
	return dfs(tree[s].lchild)&&dfs(tree[s].rchild);
	
}

int main() {
	int root = 0;
	scanf("%d", &n);
	for(int i = 0; i < n; i++) {
		string a,b;
		cin >> a >> b;
		int l = get_id(a), r = get_id(b);
		tree[i].lchild = l;
		tree[i].rchild = r;
		tree[i].index = i;
		find_farther[l]++;
		find_farther[r]++;
	}
	for(int i = 0; i < n; i++) {
		if(find_farther[i] == 0 ){
			root = i;
			break; 
		}
	}
	
	//编号
	bfs(root); 
	if(dfs(root)) {
		printf("YES %d", max_nums);
	} else {
		printf("NO %d", root);
	}
	return 0;
}