一、題目描述

原題連結

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤ 20). Then N distinct integer keys are given in the next line. All the numbers in a line are separated by a space.

​​Output Specification:

For each test case, insert the keys one by one into an initially empty AVL tree. Then first print in a line the level-order traversal sequence of the resulting AVL tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line. Then in the next line, print YES if the tree is complete, or NO if not.

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

二、解題思路

這道題目考到了AVL樹的基本操作和層序遍歷，對於AVL樹，如果出現了某個結點孩子為空，並且後面的結點還有孩子結點，則它不是一個完全二叉樹。但是AVL樹的操作還是有點複雜了，由於做的練習比較少，對AVL樹的理解也沒有很深，所以這裡也不是用的自己的程式碼，詳情可參考柳神的程式碼。
1123 Is It a Complete AVL Tree (30分)

三、AC程式碼

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
struct node {
    int val;
    struct node *left, *right;
};
node* leftRotate(node *tree) {
    node *temp = tree->right;
    tree->right = temp->left;
    temp->left = tree;
    return temp;
}
node* rightRotate(node *tree) {
    node *temp = tree->left;
    tree->left = temp->right;
    temp->right = tree;
    return temp;
}
node* leftRightRotate(node *tree) {
    tree->left = leftRotate(tree->left);
    return rightRotate(tree);
}
node* rightLeftRotate(node *tree) {
    tree->right = rightRotate(tree->right);
    return leftRotate(tree);
}
int getHeight(node *tree) {
    if (tree == NULL) return 0;
    int l = getHeight(tree->left);
    int r = getHeight(tree->right);
    return max(l, r) + 1;
}
node* insert(node *tree, int val) {
    if (tree == NULL) {
        tree = new node();
        tree->val = val;
    }else if (tree->val > val) {
        tree->left = insert(tree->left, val);
        int l = getHeight(tree->left), r = getHeight(tree->right);
        if (l - r >= 2) {
            if (val < tree->left->val)
                tree = rightRotate(tree);
            else
                tree = leftRightRotate(tree);
        }
    } else {
        tree->right = insert(tree->right, val);
        int l = getHeight(tree->left), r = getHeight(tree->right);
        if (r - l >= 2) {
            if (val > tree->right->val)
                tree = leftRotate(tree);
            else
                tree = rightLeftRotate(tree);
        }
    }
    return tree;
}
int isComplete = 1, after = 0;
vector<int> levelOrder(node *tree) {    //層序遍歷
    vector<int> v;
    queue<node *> queue;
    queue.push(tree);
    while (!queue.empty()) {
        node *temp = queue.front();
        queue.pop();
        v.push_back(temp->val);
        if (temp->left != NULL) {
            if (after) isComplete = 0;
            queue.push(temp->left);
        } else {
            after = 1;
        }
        if (temp->right != NULL) {
            if (after) isComplete = 0;
            queue.push(temp->right);
        } else {
            after = 1;
        }
    }
    return v;
}
int main() {
    int n, temp;
    scanf("%d", &n);
    node *tree = NULL;
    for (int i = 0; i < n; i++) {
        scanf("%d", &temp);
        tree = insert(tree, temp);
    }
    vector<int> v = levelOrder(tree);
    for (int i = 0; i < v.size(); i++) {
        if (i != 0) printf(" ");
        printf("%d", v[i]);
    }
    printf("\n%s", isComplete ? "YES" : "NO");
    return 0;
}