首先奉上我的平衡二叉树笔记图,丑爆+细节无序
以OJ上的这一题为例 https://acm.sdut.edu.cn/onlinejudge2/index.php/Home/Contest/contestproblem/cid/3001/pid/3374
#include<bits/stdc++.h>
using namespace std;
typedef struct treenode
{
int data;
int depth;
treenode *l,*r;
}tr;
int thedepth(tr *tn)
{
return tn == NULL? -1 : tn->depth;//注意NULL的深度是-1而不是0
}
tr* LL(tr *a)
{
tr* b = a->l;
a->l = b->r;
b->r = a;
a->depth = max(thedepth(a->l),thedepth(a->r))+1;//记得+1
b->depth = max(thedepth(b->l),thedepth(b->r))+1;
return b;
}
tr* RR(tr *a)
{
tr* b = a->r;
a->r = b->l;
b->l = a;
a->depth = max(thedepth(a->l),thedepth(a->r))+1;
b->depth = max(thedepth(b->l),thedepth(b->r))+1;
return b;
}
tr* LR(tr *a)
{
a->l = RR(a->l);//原来可以这麽巧妙
return LL(a);
}
tr* RL(tr *a)
{
a->r = LL(a->r);
return RR(a);
}
tr* plant(tr *tn,int k)//这部分框架和二叉排序树差不多,但是插入之后要判断此时树是否不平衡
{
if(tn==NULL)
{
tr *tn = new tr;
tn->data = k;
tn->depth = 0;
tn->l = NULL;
tn->r = NULL;
return tn;
}else if(k<tn->data)
{
tn->l = plant(tn->l,k);
if(thedepth(tn->l)-thedepth(tn->r)>1)//因为知道是往左下边插入的,所以不平衡只可能是左边重
{
if(k<tn->l->data)//根据情况选择对应的平衡解决方案
{
tn = LL(tn);
}else
{
tn = LR(tn);
}
}
}else
{
tn->r = plant(tn->r,k);
if(thedepth(tn->r)-thedepth(tn->l)>1)//因为知道是往右下边插入的,所以不平衡只可能是右边重
{
if(k>tn->r->data)//根据情况选择对应的平衡解决方案
{
tn = RR(tn);
}else
{
tn = RL(tn);
}
}
}
tn->depth = max(thedepth(tn->l),thedepth(tn->r))+1;//记得最后更新深度
return tn;
}
int main()
{
int n;
cin>>n;
tr* root = NULL;
for(int i =0;i<n;i++)
{
int k;
cin>>k;
root = plant(root,k);
}
cout<<root->data<<endl;
return 0;
}