【PAT】【Advanced Level】1123. Is It a Complete AVL Tree (30)

1123. Is It a Complete AVL Tree (30)

时间限制

400 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 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:

588 70 61 63 65

Sample Output 1:

70 63 88 61 65YES

Sample Input 2:

888 70 61 96 120 90 65 68

Sample Output 2:

88 65 96 61 70 90 120 68NO

原题链接：

https://www.patest.cn/contests/pat-a-practise/1123

思路：

首先平衡树

然后判断是否是满的平衡树

由于是平衡树

所以叶节点一定在最后两层

判断条件：

所有倒数第二层的叶节点在 中序遍历 序列中 一定在   层序遍历最后一个节点  的后边。

不能存在左子树为空的非叶节点

右子树为空的非叶节点一定是 层序遍历最后一个节点 的父亲

CODE：

#include<iostream>#include<vector>#include<algorithm>using namespace std;typedef struct S{int ind;int val;int ls;int rs;int ld;int rd;int fl;int tra;};vector<S> T;int trav;vector<S> ro;vector<S> le;void left(int n){S s1=T[n];S s2=T[T[n].rs];T[n].ind=n;T[n].val=s2.val;T[n].rs=s2.rs;T[n].ls=s2.ind;T[n].rd=s2.rd;T[n].ld=max(s1.ld,s2.ld)+1;T[s2.ind].ind=s2.ind;T[s2.ind].val=s1.val;T[s2.ind].rs=s2.ls;T[s2.ind].ls=s1.ls;T[s2.ind].rd=s2.ld;T[s2.ind].ld=s1.ld;}void right(int n){S s1=T[n];S s2=T[T[n].ls];T[n].ind=n;T[n].val=s2.val;T[n].ls=s2.ls;T[n].rs=s2.ind;T[n].ld=s2.ld;T[n].rd=max(s1.rd,s2.rd)+1;T[s2.ind].ind=s2.ind;T[s2.ind].val=s1.val;T[s2.ind].ls=s2.rs;T[s2.ind].rs=s1.rs;T[s2.ind].ld=s2.rd;T[s2.ind].rd=s1.rd;}int insert(int n,int v){if (v<T[n].val){if (T[n].ls==-1){S s;s.ind=T.size();s.val=v;s.ls=s.rs=-1;s.ld=s.rd=0;T[n].ls=T.size();T[n].ld=1;T.push_back(s);}else{int fl=insert(T[n].ls,v)+1;T[n].ld=fl;}}else{if (T[n].rs==-1){S s;s.ind=T.size();s.val=v;s.ls=s.rs=-1;s.ld=s.rd=0;T[n].rs=T.size();T[n].rd=1;T.push_back(s);}else{int fl=insert(T[n].rs,v)+1;T[n].rd=fl;}}if (T[n].ld-T[n].rd>1){if (T[T[n].ls].rd>T[T[n].ls].ld){left(T[n].ls);}right(n);}else if (T[n].rd-T[n].ld>1){//cout<<v<<" "<<T[n].val<<" "<<T[n].ld<<" "<<T[n].rd<<endl;if (T[T[n].rs].ld>T[T[n].rs].rd){right(T[n].rs);}left(n);}//cout<<v<<" "<<T[n].val<<" "<<T[n].ld<<" "<<T[n].rd<<endl;return max(T[n].ld,T[n].rd);}void dfs(int n,int flo){T[n].fl=flo;if (T[n].ls!=-1)dfs(T[n].ls,flo+1);T[n].tra=trav;trav++;if (T[n].rs!=-1)dfs(T[n].rs,flo+1);}bool cmp(S a, S b){if (a.fl==b.fl){return a.tra<b.tra;}else{return a.fl<b.fl;}}int main(){int n;cin>>n;S s;s.ind=0;s.ls=s.rs=-1;s.ld=s.rd=0;cin>>s.val;T.push_back(s);for (int i=1;i<n;i++){int t;cin>>t;insert(0,t);}trav=0;dfs(0,0);sort(T.begin(),T.end(),cmp);bool flag=0;for (int i=0;i<n;i++){if (i!=0) cout<<" ";cout<<T[i].val; if ( (T[i].ls==-1&&T[i].rs==-1) && (T[i].fl==T[n-1].fl-1) && (T[i].tra<T[n-1].tra)){flag=1;}if ( (T[i].ls==-1&&T[i].rs!=-1)){flag=1;}if ( (T[i].ls!=-1&&T[i].rs==-1) && (T[i].ls!=T[n-1].ind) ){flag=1;}}cout<<endl;if (flag==0)cout<<"YES";elsecout<<"NO";return 0;}