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  • 1123 Is It a Complete AVL Tree (30 分)

    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.

    F1.jpgF2.jpg
    F3.jpg F4.jpg

    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

    毕竟是第一次做平衡二叉树,

    平衡二叉树,一开始以为很难,最后看看也不过如此,我代码还是在平板上写的。

      1 #include <iostream>
      2 #include <cmath>
      3 #include <vector>
      4 #include <algorithm>
      5 #include <queue>
      6 using namespace std;
      7 struct Node {
      8   int val;
      9   Node *left, *right;
     10 } *tree;
     11 int n, m;
     12 
     13 Node *leftrotate(Node * tree) {
     14   Node *temp = tree->left;
     15   tree->left = temp->right;
     16   temp->right = tree;
     17   return temp;
     18 }
     19 
     20 Node *rightrotate(Node * tree) {
     21   Node *temp = tree->right;
     22   tree->right = temp->left;
     23   temp->left = tree;
     24   return temp;
     25 }
     26 
     27 Node *rightleftrotate(Node * tree) {
     28   tree->left = rightrotate(tree->left);
     29   return leftrotate(tree);
     30 }
     31 
     32 Node *leftrightrotate(Node * tree) {
     33   tree->right = leftrotate(tree->right);
     34   return rightrotate(tree);
     35 }
     36 
     37 
     38 int getlen(Node * root) {
     39   if (root == NULL)
     40     return 0;
     41   return max(getlen(root->left), getlen(root->right)) + 1;
     42 }
     43 
     44 Node *insert(Node * tree, int val) {
     45   if (tree == NULL) {
     46     tree = new Node();
     47     tree->val = val;
     48   } else if (val < tree->val) {
     49     tree->left = insert(tree->left, val);
     50     int ll = getlen(tree->left), rr = getlen(tree->right);
     51     if (ll - rr >= 2) {
     52       if (val < tree->left->val)
     53         tree = leftrotate(tree);
     54       else
     55         tree = rightleftrotate(tree);
     56     }
     57   } else if (val > tree->val) {
     58     tree->right = insert(tree->right, val);
     59     int ll = getlen(tree->left), rr = getlen(tree->right);
     60     if (rr - ll >= 2) {
     61       if (val > tree->right->val)
     62         tree = rightrotate(tree);
     63       else
     64         tree = leftrightrotate(tree);
     65     }
     66   }
     67   return tree;
     68 }
     69 vector<int> v, vt;
     70 void output(Node *root, int x){
     71     if(root != NULL){
     72         output(root->left, x<<1);
     73           //cout <<root->val<<endl;
     74         output(root->right, (x<<1)+1);
     75     }else{
     76         v.push_back(x);
     77     }
     78 }    
     79 
     80 void levelout(Node *root){
     81     queue<Node*> q;
     82       q.push(root);
     83     Node *ans;
     84     while(!q.empty()){
     85         ans = q.front();
     86           q.pop();
     87         vt.push_back(ans->val);
     88         if(ans->left != NULL)
     89             q.push(ans->left);
     90         if(ans->right != NULL)
     91             q.push(ans->right);
     92     }    
     93 }
     94 
     95 int main() {
     96   cin >> n;
     97   for (int i = 0; i < n; i++) {
     98     cin >> m;
     99     tree = insert(tree, m);
    100   }
    101   levelout(tree);    
    102   for(int i = 0; i < vt.size(); i++){
    103       printf("%d%c",vt[i], i == vt.size()-1?'
    ':' ');
    104   }
    105   output(tree,1);
    106   sort(v.begin(), v.end());
    107   for(int i = 1; i < v.size(); i++){
    108       if(v[i-1]+1 != v[i]){
    109         cout << "NO" << endl;
    110           return 0;
    111     }
    112   }
    113   cout << "YES" << endl;
    114   return 0;
    115 }
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  • 原文地址:https://www.cnblogs.com/zllwxm123/p/11329677.html
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