A1020 Tree Traversals

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

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 postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:
For each test case, print in one line the level order traversal sequence of the corresponding binary tree. 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:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2

#include<bits/stdc++.h> 
#define MAX 40
using namespace std;
typedef long long ll;
int in[MAX],post[MAX];
int n,k,p;
struct node
{
	int v;
	node *left;
	node *right;
	
};
//返回构建出的二叉树根节点地址 
node* create(int postl,int postr,int inl,int inr){
	if(postl>postr)
	return NULL;
	int i;
	for(i=inl;i<=inr;i++){
		if(in[i]==post[postr]) {
		break;
		}
	}
	node *t=new node();//! 
	t->v=post[postr];
	int numl=i-inl; //!
	t->left=create(postl,postl+numl-1,inl,i-1);//!
	t->right=create(postl+numl,postr-1,i+1,inr);//!
	return t; 
}
int num=0;
void bfs(node *root){
	queue<node*> que;
	que.push(root);
	while(!que.empty()){
		node *t=que.front();
		que.pop();
		cout<<t->v;
		num++;
		if(num<n) cout<<" ";
		if(t->left!=NULL) que.push(t->left);
		if(t->right!=NULL) que.push(t->right);		
	}
	
}
int main(){
	cin>>n;
	for(int i=0;i<=n-1;i++)
	cin>>post[i];
	for(int i=0;i<=n-1;i++)
	cin>>in[i];	
	
	node *root=create(0,n-1,0,n-1);
	bfs(root);
	
	
}