1123 Is It a Complete AVL Tree （PAT甲级）

这道题是看了柳婼的解法才搞定的。开始想着把height和parent放到结构体中去，很繁琐最后还搞不定……

#include <cstdio>
#include <algorithm>
#include <vector>

struct node{
    int key;
    node* left = nullptr;
    node* right = nullptr;
};

int N, t, pivot;
node* root = nullptr;
bool flag = true;
std::vector<node*> ans;

int getHeight(node* tree){
    if(!tree){
        return 0;
    }
    int l = getHeight(tree->left);
    int r = getHeight(tree->right);
    return std::max(l, r) + 1;
}

node* LL(node* tree){
    node* tmp = tree->left;
    tree->left = tmp->right;
    tmp->right = tree;
    return tmp;
}

node* RR(node* tree){
    node* tmp = tree->right;
    tree->right = tmp->left;
    tmp->left = tree;
    return tmp;
}

node* LR(node* tree){
    tree->left = RR(tree->left);
    return LL(tree);
}

node* RL(node* tree){
    tree->right = LL(tree->right);
    return RR(tree);
}

node* insertKey(node* tree, int k){
    if(!tree){
        node* tmp = new node;
        tmp->key = k;
        tree = tmp;
        return tree;
    }
    int l, r;
    if(k < tree->key){
        tree->left = insertKey(tree->left, k);
        l = getHeight(tree->left);
        r = getHeight(tree->right);
        if(l - r >= 2){
            if(k < tree->left->key){
                tree = LL(tree);
            } else{
                tree = LR(tree);
            }
        }
    } else{
        tree->right = insertKey(tree->right, k);
        l = getHeight(tree->left);
        r = getHeight(tree->right);
        if(r - l >= 2){
            if(k < tree->right->key){
                tree = RL(tree);
            } else{
                tree = RR(tree);
            }
        }
    }
    return tree;
}

int main(){
    scanf("%d", &N);
    for(int i = 0; i < N; ++i){
        scanf("%d", &t);
        root = insertKey(root, t);
    }
    ans.push_back(root);
    pivot = 0;
    while(pivot < ans.size()){
        if(ans[pivot]->left){
            ans.push_back(ans[pivot]->left);
        } else if(ans.size() != N){
            flag = false;
        }
        if(ans[pivot]->right){
            ans.push_back(ans[pivot]->right);
        } else if(ans.size() != N){
            flag = false;
        }
        ++pivot;
    }
    for(int i = 0; i < ans.size(); ++i){
        printf("%d", ans[i]->key);
        printf("%s", i == ans.size() - 1 ? "\n" : " ");
    }
    printf("%s", flag ? "YES" : "NO");
    return 0;
}

题目如下：

An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Figures 1-4 illustrate the rotation rules.

Now given a sequence of insertions, you are supposed to output the level-order traversal sequence of the resulting AVL tree, and to tell if it is a complete binary tree.

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.