PATA 1020 Tree Traversals (25 分)(利用后续和中序序列建树，输出层次遍历序列）

1020 Tree Traversals (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

题意简单清楚，注意建树时的序列划分

参考代码： 

#include <cstdio>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=50;
struct node {
	int data;
	node* lchild;
	node* rchild;
};
int pre[maxn],in [maxn], post[maxn];
int n;

//当前二叉树的后序序列区间为[postL,postR],中序序列为[inL,inR];
//create 函数返回构建出的二叉树的根节点的地址
node * create(int postL,int postR,int inL,int inR)
{
	if(postL>postR)
		return NULL; //后序序列长度小于等于0时，直接返回 
	node* root =new node;//创建根节点 
	root ->data=post[postR];
	int k;
	for(k=inL;k<=inR;++k)
	{
		if(in[k]==post[postR])//在中序序列中找到in[k]==pre[l]的结点 
			break;
	}
	int numleft=k-inL;//左子树节点个数
	
	//递归建立左子树，返回左子树的根节点地址，赋值为root所指左指针 
	root->lchild=create(postL,postL+numleft-1,inL,k-1);
	//递归建立右子树返回右子树的根节点地址，赋值为root所指右指针 
	root->rchild=create(postL+numleft,postR-1,k+1,inR);
	return root; 
}

int num=0;//已输出的结点个数
void BFS(node* root)
{
	queue<node*> q;//注意队列中存地址 
	q.push(root);
	while(!q.empty())
	{
		node* p=q.front();
		q.pop();
		printf("%d",p->data);
		num++;
		if(num<n)
			printf(" ");//访问队首元素 
		if(p->lchild)
			q.push(p->lchild); //左子树非空 
		if(p->rchild)
			q.push(p->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);//建树
	BFS(root);//层序遍历 
	return 0; 
}