# 1110. Complete Binary Tree (25)-PAT甲级真题（BFS）

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.

9
7 8
– –
– –
– –
0 1
2 3
4 5
– –
– –

YES 8

8
– –
4 5
0 6
– –
2 3
– 7
– –
– –

NO 1

#### 分析

• 判读二叉树的节点的最大下标==最大节点数量
• 使用dfs遍历出最大下标
#include

using namespace std;
struct node {
int l, r;
} a[100];
int maxn = -1, ans;

void dfs(int root, int index) {
if (index > maxn) {  //更新最大节点的id
maxn = index;
ans = root;
}
if (a[root].l != -1)//存在左孩子
dfs(a[root].l, index * 2);
if (a[root].r != -1)//存在左孩子
dfs(a[root].r, index * 2 + 1);
}

int main() {
int n, root = 0, have[100] = {0};
cin >> n;
for (int i = 0; i < n; i++) {
string l, r;
cin >> l >> r;
if (l == "-") {
a[i].l = -1;
} else {
a[i].l = stoi(l);
have[stoi(l)] = 1;
}
if (r == "-") {
a[i].r = -1;
} else {
a[i].r = stoi(r);
have[stoi(r)] = 1;
}
}
while (have[root] != 0)
root++;
dfs(root, 1);
if (maxn == n)
cout << "YES " << ans;
else
cout << "NO " << root;
return 0;
}