题目

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 tell the root of the resulting AVL tree.

Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (≤20) which is the total number of keys to be inserted. 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, print the root of the resulting AVL tree in one line.

Sample Input 1:

5
88 70 61 96 120

Sample Output 1:

70

Sample Input 2:

7
88 70 61 96 120 90 65

Sample Output 2:

88

解题思路

　　题目大意： 给个N个数，然后建立该序列的平衡二叉树，然后输出该树的根结点。
　　解题思路： 直接建树，然后输出即可（关于AVL树的讲解，详情参考）。

#include<iostream>
#include<cmath>
using namespace std;
 
struct TreeNode
{
	int value;
	TreeNode *left;
	TreeNode *right;
	int height;
	TreeNode(int v):value(v),left(NULL),right(NULL),height(0){}
	TreeNode():left(NULL),right(NULL){}
};
 
int getHeight(TreeNode *t)
{
	if(t == NULL) return -1;
	else return t->height;
}
 
int max(int a,int b) {return a>b? a:b;}
 
//LL
TreeNode *SingleRotateLeft(TreeNode *t2)
{
	TreeNode *t1;
	t1 = t2->left;
    t2->left = t1->right;
	t1->right = t2;
 
	t2->height = max(getHeight(t2->left),getHeight(t2->right)) + 1;
	t1->height = max(getHeight(t1->left),getHeight(t1->right)) + 1;
	return t1;
}
 
//RR
 TreeNode *SingleRotateRight(TreeNode *t2)
 {
	 TreeNode *t1;
	 t1 = t2->right;
	 t2->right = t1->left;
	 t1->left = t2;
 
	 t2->height = max(getHeight(t2->left),getHeight(t2->right)) + 1;
	 t1->height = max(getHeight(t1->left),getHeight(t1->right)) + 1;
	 return t1;
 }
 
 //LR
 TreeNode * DoubleRotateLR(TreeNode *t3)
 {
	 t3->left = SingleRotateRight(t3->left);
	 return SingleRotateLeft(t3);
 }
 
 //RL
 TreeNode * DoubleRotateRL(TreeNode *t3)
 {
	 t3->right = SingleRotateLeft(t3->right);
	 return SingleRotateRight(t3);
 }
 
 bool isBalanced(TreeNode *left,TreeNode *right)
 {
	 return abs(getHeight(left) - getHeight(right)) < 2;
 }
 
 TreeNode* insert(int v, TreeNode *root)
 {
	if(root == NULL)
	{
		root = new TreeNode(v);
		return root;
	}
	if(v > root->value) //节点插入在右子树中
	{
		root->right = insert(v,root->right);
		if(!isBalanced(root->left,root->right)){
			if(v > root->right->value)
				root = SingleRotateRight(root);
			else
				root = DoubleRotateRL(root);
		}
	}else{
		root->left = insert(v,root->left);
		if(!isBalanced(root->left,root->right)){
			if(v < root->left->value)
				root = SingleRotateLeft(root);
			else
				root = DoubleRotateLR(root);
		}
	}
	root->height = max(getHeight(root->left),getHeight(root->right)) + 1;
	return root;
 }
 
 int main()
 {
	 int n;
	 while(cin>>n)
	 {
		 int t;
		 TreeNode *root = NULL;
		 for(int i=0; i<n; i++)
		 {	
			 cin>>t;
			 root = insert(t,root);			 
		 }
		 cout<<root->value<<endl;
	 }
	 return 0;
 }

总结

　　这道题常规而不简单，常规常规在AVL树是书上的内容，挺常见的一个概念的 ，学过数据结构的都知道，不简单在用代码去实现AVL的人，还是比较少的。很多高级数据结构都是知其一不知其二，只会用自然语言描述，不知道怎么用代码去实现，更不知道怎么去应用。这或许就是我和大佬的差距吧，值得反思。