【PAT】【Advanced Level】1066. Root of AVL Tree (25)

1066. Root of AVL Tree (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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 ythe root of the resulting AVL tree in one line.

Sample Input 1:

588 70 61 96 120

Sample Output 1:

70

Sample Input 2:

788 70 61 96 120 90 65

Sample Output 2:

88

原题链接：

https://www.patest.cn/contests/pat-a-practise/1066

https://www.nowcoder.com/pat/5/problem/4117

思路：

比较裸的平衡树。（却写的晕头转向。。菜鸡。。）

针对四种情况，做四种不同的旋转

CODE：

#include<iostream>#define max(a, b) (((a) > (b)) ? (a) : (b))using namespace std;typedef struct S{int v;int ls,rs;int ll,rl;};S T[25];int rro(int a){int oa=a, ob=T[a].ls;int va=T[a].v;int sal=T[a].ls, sar=T[a].rs;int lal=T[a].ll, lar=T[a].rl;int vb=T[ob].v;int sbl=T[ob].ls, sbr=T[ob].rs;int lbl=T[ob].ll, lbr=T[ob].rl;{T[oa].v=vb;T[oa].ls=sbl;T[oa].rs=ob;T[oa].ll=lbl;T[oa].rl=1+max(lar,lbr);T[ob].v=va;T[ob].ls=sbr;T[ob].rs=sar;T[ob].ll=lbr;T[ob].rl=lar;}return max(T[a].ll,T[a].rl)+1;}int lro(int a){int oa=a, ob=T[a].rs;int va=T[a].v;int sal=T[a].ls, sar=T[a].rs;int lal=T[a].ll, lar=T[a].rl;int vb=T[ob].v;int sbl=T[ob].ls, sbr=T[ob].rs;int lbl=T[ob].ll, lbr=T[ob].rl;{T[oa].v=vb;T[oa].rs=sbr;T[oa].ls=ob;T[oa].rl=lbr;T[oa].ll=1+max(lal,lbl);T[ob].v=va;T[ob].ls=sal;T[ob].rs=sbl;T[ob].ll=lal;T[ob].rl=lbl;}return max(T[a].ll,T[a].rl)+1;}int ins(int a,int b){if (T[b].v>T[a].v){if (T[a].rs!=-1) T[a].rl=ins(T[a].rs,b);else{T[a].rs=b;T[a].rl=1;}}else{if (T[a].ls!=-1) T[a].ll=ins(T[a].ls,b);else{T[a].ls=b;T[a].ll=1;}}if (T[a].ll-T[a].rl>1){if (T[T[a].ls].ll>T[T[a].ls].rl){return rro(a);}else{T[a].ll=lro(T[a].ls);return rro(a);}}else if (T[a].ll-T[a].rl<-1){if (T[T[a].rs].ll>T[T[a].rs].rl){T[a].rl=rro(T[a].rs);return lro(a); }elsereturn lro(a);}elsereturn max(T[a].ll,T[a].rl)+1;}int main(){int n;cin>>n;int head=-1;for (int i=0;i<n;i++){cin>>T[i].v;T[i].ls=T[i].rs=-1;T[i].ll=T[i].rl=0;if (head==-1) head=i;else ins(head,i);}cout<<T[head].v;return 0; }