完全二叉树与数组表示法
只用一维数组，就存下整棵树

#include <iostream>
#include <cmath>
using namespace std;
int main() {
    int a[] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; // 1-based，a[0] 弃用，共 10 个结点
    int n = 10;
    cout << "i   层号  双亲  左孩子  右孩子  是否叶子" << endl;
    for (int i = 1; i <= n; i++) {
        int depth = (int)log2(i) + 1;       // 第几层
        int p  = i / 2;                       // 双亲 = ⌊i/2⌋
        int lc = 2 * i;                       // 左孩子 = 2i
        int rc = 2 * i + 1;                   // 右孩子 = 2i+1
        bool leaf = (lc > n);
        cout << i << "    " << depth << "    " << p << "    "
             << (lc <= n ? lc : 0) << "     "
             << (rc <= n ? rc : 0) << "     "
             << (leaf ? "是" : "否") << endl;
    }
    cout << "最后一个非叶子结点下标 = " << n / 2
         << "（即结点 " << a[n / 2] << "）" << endl;
    cout << "层序遍历 = 数组顺序：";
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << endl;
    return 0;
}

#include <iostream>
#include <cmath>
using namespace std;
int main() {
    int n = 10; // 0-based 编号，共 10 个结点
    cout << "i   层号  双亲  左孩子  右孩子" << endl;
    for (int i = 0; i < n; i++) {
        int depth = (int)log2(i + 1) + 1;
        int p  = (i == 0) ? -1 : (i - 1) / 2; // 双亲 = ⌊(i-1)/2⌋
        int lc = 2 * i + 1;                    // 左孩子 = 2i+1
        int rc = 2 * i + 2;                    // 右孩子 = 2i+2
        cout << i << "    " << depth << "    " << p << "    "
             << (lc < n ? lc : -1) << "     "
             << (rc < n ? rc : -1) << endl;
    }
    return 0;
}

#include <iostream>
#include <cmath>
using namespace std;
int main() {
    int n = 15;                                  // 满二叉树，15 个结点
    int k = (int)log2(n) + 1;                     // 深度 = ⌊log2 n⌋ + 1
    cout << "n = " << n
         << "  深度 = " << k
         << "  最后一个非叶子结点下标 = " << n / 2 << endl;
    return 0;
}

#include <iostream>
using namespace std;
void down(int a[], int n, int i) {           // 大顶堆向下调整（1-based）
    int largest = i;
    int l = 2 * i, r = 2 * i + 1;
    if (l <= n && a[l] > a[largest]) largest = l;
    if (r <= n && a[r] > a[largest]) largest = r;
    if (largest != i) {
        int t = a[i]; a[i] = a[largest]; a[largest] = t; // 手动交换
        down(a, n, largest);
    }
}
int main() {
    int a[] = {0, 4, 10, 3, 5, 1, 9, 7, 6, 8, 2}; // 1-based，n = 10
    int n = 10;
    cout << "建堆前：";
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << endl;
    for (int i = n / 2; i >= 1; i--) down(a, n, i); // 从最后一个非叶子 ⌊n/2⌋ 起
    cout << "建堆后（大顶堆）：";
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << endl;
    return 0;
}

#include <iostream>
#include <cmath>
using namespace std;
int main() {
    int a[] = {0, 1, 2, 3, 4, 5, 6, 7}; // 1-based，共 7 个结点（满二叉树）
    int n = 7;
    int k = (int)log2(n) + 1;
    cout << "层序遍历（= 数组顺序）：";
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << endl;
    cout << "深度 = " << k
         << "  最后一个非叶子结点下标 = " << n / 2
         << "（结点 " << a[n / 2] << "）" << endl;
    cout << "i   双亲  左孩子  右孩子" << endl;
    for (int i = 1; i <= n; i++) {
        int p = i / 2, lc = 2 * i, rc = 2 * i + 1;
        cout << i << "    " << p << "    "
             << (lc <= n ? lc : 0) << "     "
             << (rc <= n ? rc : 0) << endl;
    }
    return 0;
}

#include <iostream>
using namespace std;
void down(int a[], int n, int i) {           // 大顶堆向下调整（1-based）
    int largest = i;
    int l = 2 * i, r = 2 * i + 1;
    if (l <= n && a[l] > a[largest]) largest = l;
    if (r <= n && a[r] > a[largest]) largest = r;
    if (largest != i) {
        int t = a[i]; a[i] = a[largest]; a[largest] = t; // 手动交换
        down(a, n, largest);
    }
}
int main() {
    int a[] = {0, 4, 10, 3, 5, 1, 9, 7, 6, 8, 2}; // 1-based，n = 10
    int n = 10;
    cout << "建堆前：";
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << endl;
    for (int i = n / 2; i >= 1; i--) down(a, n, i); // 从最后一个非叶子 ⌊n/2⌋ 起
    cout << "建堆后（大顶堆）：";
    for (int i = 1; i <= n; i++) cout << a[i] << " ";
    cout << endl;
    return 0;
}