PAT_A1123#Is It a Complete AVL Tree

Source:

PAT A1123 Is It a Complete AVL Tree (30 分)

Description:

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

Keys:

Attention:

  • 判断完全二叉树，while的固定写法
  • Rotation中，先Update root，再Update temp，否则会影响结果，注意

Code:

  1 /*
  2 Data: 2019-06-24 15:36:45
  3 Problem: PAT_A1123#Is It a Complete AVL Tree
  4 AC: 35:46
  5 
  6 题目大意：
  7 由插入序列构造一棵AVL树，输出层次遍历并判断是否为一棵完全二叉树
  8 
  9 基本思路：
 10 构造平衡二叉树，
 11 中序遍历并判断是否为完全二叉树
 12 */
 13 #include<cstdio>
 14 #include<queue>
 15 #include<algorithm>
 16 using namespace std;
 17 struct node
 18 {
 19     int data;
 20     int height;
 21     node *lchild, *rchild;
 22 };
 23 
 24 int GetHeight(node *root)
 25 {
 26     if(root == NULL)
 27         return 0;
 28     else
 29         return root->height;
 30 }
 31 
 32 int GetBalanceFactor(node *root)
 33 {
 34     return GetHeight(root->lchild) - GetHeight(root->rchild);
 35 }
 36 
 37 void UpdataHeight(node *&root)
 38 {
 39     root->height = max(GetHeight(root->lchild),GetHeight(root->rchild))+1;
 40 }
 41 
 42 void LeftRotation(node *&root)
 43 {
 44     node *temp = root->rchild;
 45     root->rchild = temp->lchild;
 46     temp->lchild = root;
 47     UpdataHeight(root);
 48     UpdataHeight(temp);
 49     root = temp;
 50 }
 51 
 52 void RightRotation(node *&root)
 53 {
 54     node *temp = root->lchild;
 55     root->lchild = temp->rchild;
 56     temp->rchild = root;
 57     UpdataHeight(root);
 58     UpdataHeight(temp);
 59     root = temp;
 60 }
 61 
 62 void Insert(node *&root, int x)
 63 {
 64     if(root == NULL)
 65     {
 66         root = new node;
 67         root->data = x;
 68         root->height=1;
 69         root->lchild = root->rchild = NULL;
 70     }
 71     else if(x < root->data)
 72     {
 73         Insert(root->lchild, x);
 74         UpdataHeight(root);
 75         if(GetBalanceFactor(root) == 2)
 76         {
 77             if(GetBalanceFactor(root->lchild) == 1)
 78                 RightRotation(root);
 79             else
 80             {
 81                 LeftRotation(root->lchild);
 82                 RightRotation(root);
 83             }
 84         }
 85     }
 86     else
 87     {
 88         Insert(root->rchild, x);
 89         UpdataHeight(root);
 90         if(GetBalanceFactor(root) == -2)
 91         {
 92             if(GetBalanceFactor(root->rchild) == -1)
 93                 LeftRotation(root);
 94             else
 95             {
 96                 RightRotation(root->rchild);
 97                 LeftRotation(root);
 98             }
 99         }
100     }
101 }
102 
103 int IsComplete(node *root, int n)
104 {
105     queue<node*> q;
106     q.push(root);
107     int cnt=0, ans=1;
108     while(!q.empty())
109     {
110         root = q.front();
111         q.pop();
112         if(root)
113         {
114             printf("%d%c", root->data,++cnt==n?'\n':' ');
115             q.push(root->lchild);
116             q.push(root->rchild);
117         }
118         else
119         {
120             if(cnt==n)
121                 break;
122             else
123             {
124                 ans=0;
125                 while(!q.empty())
126                 {
127                     root = q.front();
128                     if(root) break;
129                     else   q.pop();
130                 }
131             }
132         }
133     }
134     return ans;
135 }
136 
137 
138 int main()
139 {
140 #ifdef ONLINE_JUDGE
141 #else
142     freopen("Test.txt", "r", stdin);
143 #endif // ONLINE_JUDGE
144 
145     int n,x;
146     node *root = NULL;
147     scanf("%d", &n);
148     for(int i=0; i<n; i++)
149     {
150         scanf("%d", &x);
151         Insert(root, x);
152     }
153     if(IsComplete(root, n))
154         printf("YES");
155     else
156         printf("NO");
157 
158     return 0;
159 }

 

posted @ 2019-06-03 21:14  林東雨  阅读(222)  评论(0)  编辑  收藏  举报