PAT_甲级_1123 Is It a Complete AVL Tree (30point(s)) （C++）【AVL树的构建/完全二叉树/层次遍历】

目录

​​1，题目描述​​

​​题目大意​​

​​2，思路​​

​​算法​​

​​3，AC代码​​

​​4，解题过程​​

1，题目描述

Sample Input 1:

5
88 70 61 63 65

Sample Output 1:

70 63 88 61 65
YES

Sample Input 2:

8
88 70 61 96 120 90 65 68

Sample Output 2:

88 65 96 61 70 90 120 68
NO

题目大意

根据给定序列构建AVL树，并输出AVL树的层次遍历，并判断是否为完全二叉树。

2，思路

（这一题是把以前的两道题合在一起了。。。）

​​@&再见萤火虫&【PAT_甲级_1066 Root of AVL Tree (25point(s)) (C++)【AVL树构建过程图解】】​​

​​@&再见萤火虫&【PAT_甲级_1110 Complete Binary Tree (25point(s)) (C++)【完全二叉树】】​​

算法

1，构建AVL树：

LL/RR/LR/RL型旋转：

计算树的高度

向AVL树中插入节点

2，利用队列层次遍历AVL树 ：

3，利用DFS计算节点的最远距离maxDep（即将树放在下标从1开始的数组中，下标可达的最远位置，也就是最后一个节点的位置，若maxDep大于N，说明数组中间有部分节点未填充）判断是否为完全二叉树；

3，AC代码

#include<bits/stdc++.h>
using namespace std;
int N, maxDep = 0;      //maxDep判断是否为完全二叉树
vector<int> level;
struct node{
    int key;
    node *left = NULL, *right = NULL;
};
int getHeight(node *tree){
    if(tree == NULL)
        return 0;
    return max(getHeight(tree->left), getHeight(tree->right)) + 1;
}
node *LL(node *root){   //不需要node *root
    node *tem = root->left;
    root->left = tem->right;
    tem->right = root;
    return tem;
}
node *RR(node *root){
    node *tem = root->right;
    root->right = tem->left;
    tem->left = root;
    return tem;
}
node *LR(node *root){
    node *A = root, *B = root->left, *C = root->left->right;// !!!
    A->left = C->right;
    B->right = C->left;
    C->left = B;C->right = A;
    return C;
}
node *RL(node *root){
    node *A = root, *B = root->right, *C = root->right->left;// !!!
    A->right = C->left;
    B->left = C->right;
    C->left = A;C->right = B;
    return C;
}
node *insertAVL(node *root, int key){   //不需要node *&root
    if(root == NULL){
        root = new node();
        root->key = key;
        root->left = root->right = NULL;
        return root;
    }
    else if(key >= root->key){
        root->right = insertAVL(root->right, key);  // !!!不是root = insertAVL(root->right, key);
        if(getHeight(root->right) - getHeight(root->left) > 1){
            if(key >= root->right->key)
                root = RR(root);
            else
                root = RL(root);
        }

    }else{
        root->left = insertAVL(root->left, key);    // !!!不是root = insertAVL(root->left, key);
        if(getHeight(root->left) - getHeight(root->right) > 1){
            if(key < root->left->key)
                root = LL(root);
            else
                root = LR(root);
        }
    }
    return root;
}
void levelOrder(node *root){            //用于获得层次遍历
    if(root == NULL)
        return;
    queue<node *> q;
    q.push(root);

    while(!q.empty()){
        node *tem = q.front();
        level.push_back(tem->key);
        q.pop();
        if(tem->left != NULL) q.push(tem->left);
        if(tem->right != NULL) q.push(tem->right);
    }
}
void dfs(node *root, int dep){          //用于判断是否为完全二叉树
    if(root == NULL) return;// !!!
    if(root->left == NULL && root->right == NULL){
        maxDep = max(maxDep, dep);
        return;
    }
    dfs(root->left, 2 * dep);
    dfs(root->right, 2 * dep + 1);
}
int main(){
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif // ONLINE_JUDGE
    cin>>N;
    int num;
    node *root = NULL;
    for(int i = 0; i < N; i++){
        cin>>num;
        root = insertAVL(root, num);
    }
    levelOrder(root);

    for(int i = 0; i < N; i++)
        printf("%d%c", level[i], i == N-1 ? '\n':' ');
    dfs(root, 1);
    printf("%s", maxDep == N ? "YES":"NO"); //根据完全二叉树的性质 maxDep==N时即为完全二叉树
    return 0;
}

4，解题过程

测试用例期间有些BUG，不过测试一次通过o(*￣▽￣*)ブ