1020 Tree Traversals (25)（25 分）

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<cstdio>
#include<cstring>
#include<queue>
#include<fstream>
#include<algorithm>
using namespace std;
const int maxn=50;

struct node{
	int data;
	node* lchild;
	node* rchild;
};
int in[maxn], post[maxn];
int n;

node* create(int postl, int postr, int inl, int inr){
	if(postl>postr){
		return NULL;
	}
	node* root=new node;
	root->data=post[postr];
	int k;
	for(k=inl; k<=inr; k++){
		if(in[k]==post[postr]){
			break;
		}
	}
	int numleft=k-inl;
	root->lchild=create(postl, postl+numleft-1, inl, k-1);
	root->rchild=create(postl+numleft, postr-1, k+1, inr);
	return root;
}

int num=0;
void BFS(node* root){
	queue<node*> q;
	q.push(root);
	while(!q.empty()){
		node* now=q.front();
		q.pop();
		printf("%d", now->data);
		num++;
		if(num<n) printf(" ");
		if(now->lchild!=NULL) q.push(now->lchild);
		if(now->rchild!=NULL) q.push(now->rchild);  
	} 
}

int main(){
//	freopen("d://in.txt","r",stdin);
	scanf("%d", &n);
	for(int i=0; i<n; i++){
		scanf("%d", &post[i]);
	}
	for(int i=0; i<n; i++){
		scanf("%d", &in[i]);
	}
	node* root=create(0, n-1, 0, n-1);
	BFS(root);
	return 0;
}