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

题目大意：给出一个n表示有n个结点，这n个结点为0~n-1，给出这n个结点的左右孩子，求问这棵树是不是完全二叉树
分析：递归出最大的下标值，完全二叉树一定把前面的下标充满：最大的下标值 == 最大的节点数；不完全二叉树前满一定有位置是空，会往后挤：最大的下标值 > 最大的节点数～

#include <iostream>
using namespace std;
struct node{
    int l, r;
}a[100];
int maxn = -1, ans;
void dfs(int root, int index) {
    if(index > maxn) {
        maxn = index;
        ans = root;
    }
    if(a[root].l != -1) dfs(a[root].l, index * 2);
    if(a[root].r != -1) 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;
}

#include <iostream>

using namespace std;

struct node{

int l, r;

}a[100];

int maxn = -1, ans;

void dfs(int root, int index) {

if(index > maxn) {

maxn = index;

ans = root;

}

if(a[root].l != -1) dfs(a[root].l, index * 2);

if(a[root].r != -1) 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;

}

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