二叉树 | 静态实现

晴神笔记十分贴心呐，我就是那种对指针写法信心不足的读者。所以下面来看看数组实现二叉树吧～

静态二叉链表

• 将struct Node的指针域用int代替，表示左右子树根结点在数组中的下标。
• 由于需要用到下标，要预先开一个大小为结点上限个数的Node型数组。
• 以下标为“ -1”表示孩子结点为空。

例

1102 Invert a Binary Tree (25 分)

⚠️%c格式化输入时，getchar（）吞掉前面的换行符，或者scanf("%*c%c %c", lc, rc);也可，getchar()、%*c这两种写法可以吞掉任意一个字符。
⚠️此题要求反转左右子树，建树时反转最方便。
⚠️找出根：不是任何结点的孩子结点的结点就是根。

#include <cstdio>
#include <algorithm>
#include <queue>

#define MAXN 10
using namespace std;

int nn, cnt1, cnt2;
bool ancestor[MAXN];
struct Node {
    int lchild = -1, rchild = -1;
} btree[MAXN];

void level_order(int root) {
    queue<int> mq;
    mq.push(root);
    while (!mq.empty()) {
        int curr = mq.front();
        printf("%d", curr);
        cnt1++;
        if (cnt1 < nn) printf(" ");
        mq.pop();
        if (btree[curr].lchild != -1)
            mq.push(btree[curr].lchild);
        if (btree[curr].rchild != -1)
            mq.push(btree[curr].rchild);
    }
}

void in_order(int root) {
    if (btree[root].lchild != -1) {
        in_order(btree[root].lchild);
    }
    cnt2++;
    printf("%d", root);
    if (cnt2 < nn) printf(" ");
    if (btree[root].rchild != -1) {
        in_order(btree[root].rchild);
    }
}

int main() {
    fill(ancestor, ancestor + MAXN, true);
    scanf("%d", &nn);
    char lc, rc;
    for (int i = 0; i < nn; ++i) {
        getchar();
        scanf("%c %c", &lc, &rc);
        if (lc != '-') {
            btree[i].rchild = lc - '0';
            ancestor[lc - '0'] = false;
        }
        if (rc != '-') {
            btree[i].lchild = rc - '0';
            ancestor[rc - '0'] = false;
        }
    }
    int root;
    for (int i = 0; i < nn; ++i) {
        if (ancestor[i]) {
            root = i;
            break;
        }
    }
    cnt1 = cnt2 = 0;
    level_order(root);
    puts("");
    in_order(root);
    return 0;
}

1099 Build A Binary Search Tree (30 分)

#include <cstdio>
#include <algorithm>
#include <queue>

using namespace std;
struct Node {
    int value;
    int lchild, rchild;
} b_tree[101];
int nn, seq[101], temp;

void in_order_traverse(int root) {
    if (b_tree[root].lchild != -1) in_order_traverse(b_tree[root].lchild);
    b_tree[root].value = seq[temp++];
    if (b_tree[root].rchild != -1) in_order_traverse(b_tree[root].rchild);
}

void level_order_traverse(int root) {
    queue<int> mq;
    mq.push(root);
    while (!mq.empty()) {
        int curr = mq.front();
        mq.pop();
        printf("%d", b_tree[curr].value);
        temp--;
        if (temp > 0) printf(" ");
        if (b_tree[curr].lchild != -1) mq.push(b_tree[curr].lchild);
        if (b_tree[curr].rchild != -1) mq.push(b_tree[curr].rchild);
    }
}

int main() {
    scanf("%d", &nn);
    for (int i = 0; i < nn; ++i)
        scanf("%d%d", &b_tree[i].lchild, &b_tree[i].rchild);
    for (int i = 0; i < nn; ++i)
        scanf("%d", &seq[i]);
    sort(seq, seq + nn);
    temp = 0;
    in_order_traverse(0);
    level_order_traverse(0); //BFS
    return 0;
}