1020 Tree Traversals -PAT甲级

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

简单的二叉树的重建与遍历，利用后序序列与中序序列进行二叉树的重建，利用层次遍历来输出最后的结果

套用模板即可

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
const int N=35;
int post[N],in[N];
int n;
struct node{
	int data;
	node *lchild=NULL;
	node *rchild=NULL;
};
node *create(int postl,int postr,int inl,int inr){
	if(postl>postr){
		return NULL;
	}
	node *root=new node;//新建一个结点 
	root->data=post[postr];
	int k;
	for(k=inl;k<=inr;k++){
		if(in[k]==post[postr])//在中序遍历中找到对应的根节点的位置 
			break;
	}
	int numleft=k-inl;//左子树的结点的个数 
	root->lchild=create(postl,postl+numleft-1,inl,k-1);
	root->rchild=create(postl+numleft,postr-1,k+1,inr);
	return root;
}
void layer_order(node *root){
	queue<node *>q;//队列里面存的是结点的地址
	q.push(root);
	int flag=0;
	while(!q.empty()){
		node *now=q.front();
		q.pop();
		if(flag) cout<<" ";
		flag=1;
		cout<<now->data;
		if(now->lchild!=NULL){
			q.push(now->lchild);
		}
		if(now->rchild!=NULL){
			q.push(now->rchild);
		}
	} 
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>post[i];
	for(int i=1;i<=n;i++)
		cin>>in[i];
	node *root=create(1,n,1,n);
	layer_order(root);
	return 0;
}