【AVL树】PAT甲级 1123 Is It a Complete AVL Tree

1123 Is It a Complete AVL Tree

看柳神的代码，非常之简洁，推荐~

解题思路

LL 型的时候：对根节点使用一次右旋（代码中即函数 R()，意思是进行右旋）

LR 型的时候：对根节点的左节点先进行左旋，再对根节点进行右旋（代码中即函数 LR()，和形状命名相同）

RR、RL 型同理。

参考代码

using namespace std;
typedef long long LL;
typedef double db;
const int inf = 0x3f3f3f3f;
const LL INF = 1e18;
const int N = 200 + 10;

inline int readint() {int x; scanf("%d", &x); return x;}

struct node{
    int val;
    node *l, *r;
    node() {l = r = NULL;}
};

node* L(node *tree) {  //左旋：RR的时候用
    node *temp = tree->r;
    tree->r = temp->l;
    temp->l = tree;
    return temp;
}

node *R(node *tree) {  //右旋：LL的时候用
    node *temp = tree->l;
    tree->l = temp->r;
    temp->r = tree;
    return temp;
}

node *LR(node *tree) {
    tree->l = L(tree->l);
    return R(tree);
}

node *RL(node *tree) {
    tree->r = R(tree->r);
    return L(tree);
}

int get_h(node *tree) {
    if (tree == NULL) return 0;
    int l = get_h(tree->l);
    int r = get_h(tree->r);
    return max(l, r) + 1;
}

node *insert(node *tree, int val) {
    if (tree == NULL) {
        tree = new node;
        tree->val = val;
    }
    else if (val < tree->val) {
        tree->l = insert(tree->l, val);
        int l = get_h(tree->l), r = get_h(tree->r);
        if (l - r >= 2) {  //LL或LR型
            if (val < tree->l->val) {  //LL
                tree = R(tree);
            }
            else  //LR
                tree = LR(tree);
        }
    }
    else {
        tree->r = insert(tree->r, val);
        int l = get_h(tree->l), r = get_h(tree->r);
        if (r - l >= 2) {  //RR或RL型
            if (val > tree->r->val)  //RR
                tree = L(tree);
            else  //RL
                tree = RL(tree);
        }
    }
    return tree;
}

int is_complete = 1, after = 0;  //after用于判断当前结点之后是不是都要NULL，为1时代表是
VI level_order(node *tree) {
    VI ans;
    queue<node*> q;
    q.push(tree);
    while (!q.empty()) {
        node *now = q.front();
        q.pop();
        ans.push_back(now->val);
        if (now->l != NULL) {
            if (after) {
                is_complete = 0;
            }
            q.push(now->l);
        }
        else after = 1;
        if (now->r != NULL) {
            if (after) is_complete = 0;
            q.push(now->r);
        }
        else after = 1;
    }
    return ans;
}
int main() {
#ifdef LOCAL
    freopen("input.txt", "r", stdin);
//   freopen("output.txt", "w", stdout);
#endif
    int n = readint();
    node *root = NULL;
    for(int i = 0; i < n; ++i) {
        int val = readint();
        root = insert(root, val);
    }
    VI ans;
    ans = level_order(root);
    for(int i = 0; i < (int)ans.size(); ++i) {
        printf("%s%d", i ? " " : "", ans[i]);
    }
    printf("\n%s", is_complete ? "YES" : "NO");

    return 0;
}