PAT 甲级 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

题意：给出该二叉树有多少个结点，后面的第一个序列是后序遍历序列，第二个是中序遍历序列，根据这两个序列求层次遍历序列。

思路:根据两个遍历序列构造二叉树，然后用层序遍历（广度优先搜索--分岔口总是先访问可以直接到达的结点（将结点的左右孩子先放入队列内））求出层次遍历序列

代码：
 

#include<iostream>
#include<queue>
using namespace std;
const int maxn=50;
int in[maxn],post[maxn];
int n;
struct node{
	int data;
	node* lchild;
	node* rchild;
};
//根据序列构造二叉树; 
node* create(int postL,int postR,int inL,int inR){
	if(postL>postR)return NULL;//递归边界;
	node* root=new node;//构造根节点; 
	root->data=post[postR];
	int k;
	for(k=inL;k<=inR;k++){//注意这里是inL到inR，在子树内寻找跟根节点,不是从0到inR，还有因为传入的是0到n-1，所以是可以等于inR的; 
		if(in[k]==root->data){
			break;
		}
	}
	int leftnum=k-inL;//确认左右子树;
	root->lchild=create(postL,postL+leftnum-1,inL,k-1);//然后进入左子树重复该操作，将两个序列带入函数，找到根节点，创建根节点，确认左右子树，直到子序列没数时（为空树的时候），返回NULL； 
	root->rchild=create(postL+leftnum,postR-1,k+1,inR);//递归式； 
	return root; 
}
//广度优先搜索求层次序列; 
void BFS(node* root){
	queue<node* > Q; 
	//注意这里是指针队列,因为队列内存放的是二叉树结点的副本，如果想要对二叉树的数据进行修改，需要用到该结点的指针，去到该结点的位置进行修改，因此这存放的是指针; 
	Q.push(root);
	int num=0;
	while(!Q.empty()){
		node* top=Q.front();
		Q.pop();
		printf("%d",top->data);
		num++;
		if(num<n)printf(" ");
		if(top->lchild!=NULL)Q.push(top->lchild);
		if(top->rchild!=NULL)Q.push(top->rchild);
	}
}
int main( ){
	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;
}