PAT 1020 Tree Traversals binary tree traversal

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

#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
int in[50],post[50];
struct node{
	int val;
	node *lchild;
	node *rchild;
};
int n;
node* create(int in_l,int in_r,int post_l,int post_r){
	
	if(post_r<post_l)
		return NULL;
	node *root = new node;
	root->val=post[post_r];
	//理解distinct含义:n个有区别的、不同的数字
	//后序遍历 当前子树的最后一个结点是根，传入中序遍历中，划分得到左右子树 
	int k;
	for(k=in_l;k<=in_r;k++)
		if(in[k]==post[post_r])
			break; 
	int numleft=k-in_l;
	root->lchild=create(in_l,k-1,post_l,post_l+numleft-1);
	root->rchild=create(k+1,in_r,post_l+numleft,post_r-1);
	return root;
}
int cnt=0;
void BFS(node *root){
	queue<node*>q;
	q.push(root);
	while(!q.empty()){
		node *now=q.front();
		q.pop();
		printf("%d",now->val);
		cnt++;
		if(cnt<n) printf(" ");
		if(now->lchild != NULL) q.push(now->lchild);
		if(now->rchild != NULL) q.push(now->rchild);
	}
}
int main(){
	scanf("%d",&n);
	//后序 
	for(int i=0;i<n;i++)
		scanf("%d",&post[i]);
	//中序 
	for(int i=0;i<n;i++)
		scanf("%d",&in[i]);
	node *root = create(0,n-1,0,n-1);
	BFS(root);
	return 0;
}