PAT 1123 Is It a Complete AVL Tree (30分)（完全二叉平衡树）

1123 Is It a Complete AVL Tree (30分)

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.

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：判断是否是完全二叉平衡树（完全二叉树+AVL树）

1、判断是不是完全二叉树，就看出现一个孩子为空的结点之后是否还会出现孩子为空的情况，是的话就不是完全二叉树，否则是

2、AVL树，建树，发现树不平衡然后根据旋转规则来旋转

A发现树不平衡的情况有四种：
新来的结点插入到A的左子树的左子树，解决：右旋
新来的结点插入到A的左子树的右子树，解决：左右旋
新来的结点插入到A的右子树的左子树，解决：右左旋
新来的结点插入到A的右子树的右子树，解决：左旋

（真是个好题，学到很多）

//给你插入的序列，让你输出二叉平衡树的层序序列，然后判断是不是完全二叉树 
#include <bits/stdc++.h>
using namespace std;
struct node{
  int val;
  node *left, *right;
  node(int v):val(v), left(NULL), right(NULL){}
}; 
int n, isComplete, after;
vector<int> ans;

//左旋 
node* leftRotate(node *rt) {
  node* temp = rt->right;
  rt->right = temp->left;
  temp->left = rt;
  return temp; 
}
//右旋 
node* rightRotate(node *rt) {
  node* temp = rt->left;
  rt->left = temp->right;
  temp->right = rt;
  return temp; 
} 
//左右旋 
node* leftRightRotate(node *rt) {
  rt->left = leftRotate(rt->left);
  return rightRotate(rt); 
}
//右左旋 
node* rightLeftRotate(node *rt) {
  rt->right = rightRotate(rt->right);
  return leftRotate(rt);
}
int getHeight(node *rt) {
  if(rt == NULL) return 0;
  int l = getHeight(rt->left);
  int r = getHeight(rt->right);
  return max(l, r) + 1;
}
node* insert(node *rt, int val) {
  if(rt == NULL) {
    rt = new node(val);
  }else if(val < rt->val) {  //左边插 
    rt->left = insert(rt->left, val);
    int l = getHeight(rt->left);
    int r = getHeight(rt->right);
    if(l - r >= 2) {
      if( val < rt->left->val) { //左左情况   
        rt = rightRotate(rt); //右旋 
      }else {  //左右情况 
        rt = leftRightRotate(rt); //左右旋 
      }
    }
  }else {
    rt->right = insert(rt->right, val);
    int l = getHeight(rt->left);
    int r = getHeight(rt->right);
    if(r - l >= 2) {
      if(val > rt->right->val ){  //右右情况 
        rt = leftRotate(rt); //左旋 
      }else {
        rt = rightLeftRotate(rt);  //右左情况  右左旋 
      }
    }
  }
  return rt;
}
void levelOrder(node* rt){
  queue<node *> q;
  q.push(rt);
  while(!q.empty()) {
    node *temp = q.front();
    q.pop();
    ans.push_back(temp->val);
    if(temp->left != NULL) {
      if(after) isComplete = 0;
      q.push(temp->left);
    } else {
      after = 1;
    }
    if(temp->right != NULL) {
      if(after) isComplete = 0;
      q.push(temp->right);
    }else after = 1;
  }
}
int main() {
  int temp;
  node *root = NULL;
  ans.clear(); isComplete = 1, after = 0;
  cin >> n;
  for(int i = 0; i < n; i++) {
    cin >> temp;
    root = insert(root, temp);
  }
  levelOrder(root);
  for(int i = 0; i <ans.size(); i++) {
    cout << ans[i];
    if(i != ans.size() - 1) cout << " ";
    else cout << endl;
  }
  if(isComplete) cout << "YES" << endl;
  else cout << "NO" << endl; 
  return 0;
}