已知下列字符 A、B、C、D、E、F、G 的权值分别为 3、12、7、4、2、8，11，试 填写出其对应哈夫曼树 HT 的存储结构的初态和终态

初态：

struct HTNode {
    int weight;
    int parent;
    int leftChild;
    int rightChild;
};
HTNode HT[2 * n - 1]; // n为叶子节点数目，即7

// 初始化叶子节点
for (int i = 0; i < n; i++) {
    HT[i].weight = w[i];
    HT[i].parent = -1;
    HT[i].leftChild = -1;
    HT[i].rightChild = -1;
}

终态：

// 构造哈夫曼树
for (int i = n; i < 2 * n - 1; i++) {
    HT[i].weight = 0;
    HT[i].parent = -1;
    HT[i].leftChild = -1;
    HT[i].rightChild = -1;
}

for (int i = n; i < 2 * n - 1; i++) {
    int min1 = INT_MAX, min2 = INT_MAX;
    int x1 = -1, x2 = -1;
    for (int j = 0; j < i; j++) {
        if (HT[j].parent == -1) {
            if (HT[j].weight < min1) {
                min2 = min1;
                x2 = x1;
                min1 = HT[j].weight;
                x1 = j;
            } else if (HT[j].weight < min2) {
                min2 = HT[j].weight;
                x2 = j;
            }
        }
    }
    HT[x1].parent = i;
    HT[x2].parent = i;
    HT[i].leftChild = x1;
    HT[i].rightChild = x2;
    HT[i].weight = HT[x1].weight + HT[x2].weight;
}