Play with binary tree (25 points)

Given a binary tree in order to traverse and preorder traversal , Please mirror the tree first , And then output the inverted sequence traversal sequence . The so-called mirror reversal , It means to exchange the left and right children of all non leaf nodes . It is assumed that the key values are all positive integers which are not equal to each other .

Input format ：

The first line of input gives a positive integer N（≤30）, Is the number of nodes in a binary tree . The second line gives the middle order traversal sequence . The third line gives its preorder traversal sequence . Numbers are separated by spaces .

Output format ：

Output the sequence traversed by the inverted sequence of the tree in one line . Between the numbers 1 Space separation , There must be no extra space at the beginning and end of the line .

sample input ：

7
1 2 3 4 5 6 7
4 1 3 2 6 5 7

sample output ：

4 6 1 7 5 3 2

#include<bits/stdc++.h>
using namespace std;
const int N=40;
map<int,int>l,r,pos;
int post[N],in[N];
int n,cnt;
int build(int il,int ir,int pl,int pr){
	int root=post[pl];
	int k=pos[root];
	if(il<k)l[root]=build(il,k-1,pl+1,pl+k-il);
	if(ir>k)r[root]=build(k+1,ir,pl+k-il+1,pr);
	return root;
}
void bfs(int root)                          //BFS Used to traverse the output 
{
	queue<int> q;
	q.push(root);
	while (q.size())
	{
		auto t = q.front();
		q.pop();
		if(cnt==0)cout<<t;
		else cout <<" "<<t;
		cnt++;
		if (r.count(t)) q.push(r[t]);       // If it exists, join the queue , Wait for the next layer to traverse 
		if (l.count(t)) q.push(l[t]);       // Judge whether the left and right sons of the node exist 

	}
}
int main(){
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>in[i];
		pos[in[i]]=i;
	}
	for(int i=0;i<n;i++)cin>>post[i];
	int root=build(0,n-1,0,n-1);
	bfs(root);
	return 0;
}