1110 Complete Binary Tree

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

判断是否是完全二叉树：首先树的高度和结点个数满足 n >= power(2, h - 1) && n <= (power(2, h) - 1)，其次最下面一层的结点必须是从左到右依次排列的，即倒数第二层的结点不能只有右子树而没有左子树，可以只有左子树，而且如果序号在前的结点已经没有左右子树了，序号在后的结点就不能再有左右子树。

#include 
#include 
#include 
#include 
#include 
using namespace std;
int n, root, cnt, last, h;
string l, r;
bool vis[50];
bool flag;
 
struct node {
	int val, left, right;
} a[50];
queueq;
 
int power(int x, int y) {
	int ans = 1;
	while (y > 0) {
		if (y & 1) {
			ans *= x;
		}
		x *= x;
		y >>= 1;
	}
	return ans;
}
 
void dfs(int x, int d) {
	h = max(h, d);
	if (a[x].left != -1) {
		dfs(a[x].left, d + 1);
	}
	if (a[x].right != -1) {
		dfs(a[x].right, d + 1);
	}
}
 
int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> l >> r;
		a[i].val = i;
		if (l == "-") {
			a[i].left = -1;
		} else {
			a[i].left = atoi(l.c_str());
			vis[atoi(l.c_str())] = 1;
		}
		if (r == "-") {
			a[i].right = -1;
		} else {
			a[i].right = atoi(r.c_str());
			vis[atoi(r.c_str())] = 1;
		}
		if (a[i].left == -1 && a[i].right != -1) {
			flag = 1;
		}
	}
	for (int i = 0; i < n; i++) {
		if (!vis[i]) {
			root = i;
			break;
		}
	}
	if (flag) {
		cout << "NO " << root;
		return 0;
	}
	dfs(root, 1);
	if (!(n >= power(2, h - 1) && n <= (power(2, h) - 1))) {
		cout << "NO " << root;
		return 0;
	}
	q.push(a[root]);
	bool child = 1;
	while (!q.empty()) {
		int u = q.front().val;
		cnt++;
		q.pop();
		if (a[u].left != -1) {
			q.push(a[a[u].left]);
		}
		if (a[u].right != -1) {
			q.push(a[a[u].right]);
		}
		if (a[u].left != -1 && !child) {
			flag = 1;
		}
		if (a[u].right == -1 && child) {
			child = 0;
		}
		if (cnt == n) {
			last = u;
		}
	}
	if (flag) {
		cout << "NO " << root;
	} else {
		cout << "YES " << last;
	}
	return 0;
}