1020 Tree Traversals （25 分)

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:知道二叉树的后序和中序遍历，求二叉树的层序遍历。（类似的题目有知道二叉树前序和中序，重建这棵二叉树）。

注意:左右子树递归的时的参数即可，最好画图。

#include 
#include 
#include 
using namespace std;

const int maxn=50;
struct node{
	int data;
	node *l;
	node *r;
};

int post[maxn],in[maxn],n;
node *create(int inL,int inR,int postL,int postR)//重建一棵树 
{
	if(postL>postR){
		return NULL;
	}
	node *root=new node;//新建结点，用于存放根 
	root->data=post[postR];//为根结点赋值 
	int k;//记录中序的根在哪个位置 
	for(k=inL;k<=inR;k++){
		if(in[k]==post[postR]){
			break;
		}
	}
	int numLeft=k-inL;//左子树的结点个数 
	root->l=create(inL,k-1,postL,postL+numLeft-1);
	root->r=create(k+1,inR,postL+numLeft,postR-1);

	return root;
}
int num=0;
void laywerOrder(node *root)
{
	queue q;
	q.push(root);
	while(!q.empty()){
		node *now=q.front();
		q.pop();
		printf("%d",now->data);
		num++;
		if(numl!=NULL){
			q.push(now->l);
		}
		if(now->r!=NULL){
			q.push(now->r);
		}
	}
}

int main()
{
	scanf("%d",&n);
	
	for(int i=0;i