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.

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

4 1 6 3 5 7 2

AC代码：

//后序遍历和中序遍历序列能唯一确定一棵二叉树

#include<stdio.h>
#include<queue>
using namespace std;

struct Node {
    Node *lchild;
    Node *rchild;
    int num;
};

int loc; //存储节点个数
int str1[30],str2[30];

void levelOrder(Node *root) { //利用STL的队列进行层序遍历
	queue<Node *> q;
	if(root)
		q.push(root); //根节点入队列
	while(!q.empty()) {
		Node *t = q.front(); //取得队头元素
		q.pop(); //弹出队头元素
		if(loc!=1)
			printf("%d ",t->num);
		else 
			printf("%d",t->num); //最后一个节点后不输出空格
		loc--;
		if(t->lchild) { //若左孩字存在，则入队列
			q.push(t->lchild);
		}
		if(t->rchild) { //若右孩字存在，则入队列
			q.push(t->rchild);
		}
	}
}

Node *build(int s1, int e1, int s2, int e2) {
	Node *ret = new Node();	//申请一个新的节点
	ret->lchild = ret->rchild = NULL; //初始化左右孩子为空
    ret->num = str1[e1]; //将后序最后一个节点赋给当前节点
    int rootldx;
    for(int i=s2; i<=e2; i++) {
        if(str2[i] == str1[e1]) {
            rootldx = i; //寻找当前根节点在中序序列中的位置
            break;
        }            
    }
    if(rootldx != s2) { //中序遍历：左-根-右，当根节点位于序列头时表明左孩子为空
        ret->lchild = build(s1,s1 + (rootldx-s2) - 1,s2,rootldx-1);
    }
    if(rootldx != e2) {
        ret->rchild = build(s1 + (rootldx-s2),e1-1,rootldx+1,e2);
    }
    return ret;
}

int main()
{
    //freopen("in.txt","r",stdin);
    int n;
    while(scanf("%d",&n) != EOF) {
        for(int j=0; j<n; j++) {
            scanf("%d ",&str1[j]);
        }
        for(int k=0; k<n; k++) {
            scanf("%d",&str2[k]);
        }
        loc = n;
        Node *root = build(0,n-1,0,n-1);
        levelOrder(root);
		printf("\n");
    }
    return 0;
}