pat 甲1127. ZigZagging on a Tree (已知后序及中序建树，并层次往返输出)

1127. ZigZagging on a 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

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However,
if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in "zigzagging order" -- that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left.
For example, for the following tree you must output: 1 11 5 8 17 12 20 15.



Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<= 30), the total number of nodes in the binary tree. The second line gives the inorder sequence and the third line gives the postorder sequence. All the numbers
in a line are separated by a space.

Output Specification:

For each test case, print the zigzagging sequence of the tree in a line. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:

8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1

Sample Output:

1 11 5 8 17 12 20 15

#include<iostream>
#include<queue>
#include<vector>
using namespace std;

int n,rt,maxn;//节点数量，根，树的最大深度
int in[55],post[55];
int t[35][2];
vector<int>ans[55];//存放每一层的结点信息

void dfs(int &x,int l,int r,int ll,int rr)
{
if(l>r)return;
x=rr;//存放的是下标

for(int i=l;i<=r;i++)
{
if(in[i]==post[rr])
{
dfs(t[x][0],l,i-1,ll,ll+i-1-l);
dfs(t[x][1],i+1,r,ll+i-l,rr-1);
}
}

}
struct node{
int index,depth;
};
void bfs()
{
queue<node>q;
q.push(node{rt,0});
while(!q.empty())
{
node tt=q.front();q.pop();
maxn=max(maxn,tt.depth);
ans[tt.depth].push_back(tt.index);
if(t[tt.index][0])q.push({t[tt.index][0],tt.depth+1});
if(t[tt.index][1])q.push({t[tt.index][1],tt.depth+1});
}
}
int flag;
void output()
{
for(int deep=0;deep<=maxn;deep++)
{
if(deep&1)
for(int i=0;i<ans[deep].size();i++){
if(!flag)cout<<post[ans[deep][i]];
else cout<<" "<<post[ans[deep][i]];
flag=1;
}
else
for(int i=ans[deep].size()-1;i>=0;i--){
if(!flag)cout<<post[ans[deep][i]];
else cout<<" "<<post[ans[deep][i]];
flag=1;
}
}
}
int main()
{
cin>>n;
for(int i=1;i<=n;i++)cin>>in[i];
for(int i=1;i<=n;i++)cin>>post[i];

dfs(rt,1,n,1,n);//建树
bfs();//存储每层节点
output();//按照规则输出
putchar(10);
return 0;
}