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.

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

题目理解

给出一棵树，判断这棵树是否是一个完全二叉树
如果是完全二叉树，输出“YES”和最后一个节点的id
如果不是，输出“NO”和根节点root

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;
}