个人学习记录，代码难免不尽人意。

Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in “zigzagging order” – that is, starting from the root, print the numbers level-by-level, alternating between left to right and right to left. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.

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 inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.

Output Specification:
For each test case, print the zigzagging sequence of the tree in a line. 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:
8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1
Sample Output:
1 11 5 8 17 12 20 15

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cmath> 
#include<queue>
#include<string>
using namespace std;
const int maxn=35;
int inorder[maxn];
int postorder[maxn];
struct node{
	int data;
	node* lchild;
	node* rchild;
};
node* newnode(int data){
	node* root=new node;
	root->data=data;
	root->lchild=NULL;
	root->rchild=NULL;
	return root;
}
node* create(int postl,int postr,int inl,int inr){
	if(postl>postr) return NULL;
	int head=postorder[postr];
	node* root=newnode(head);
	int index;
	for(int i=inl;i<=inr;i++){
		if(inorder[i]==head){
			index=i;
			break;
		}
	}
	int left=index-inl;
	root->lchild=create(postl,postl+left-1,inl,inl+left);
	root->rchild=create(postl+left,postr-1,index+1,inr);
	return root;
}
int zorder[maxn];
void BFS(node* root,int n){
	int num=0;
	bool flag=true;
	queue<node*> q;
	vector<node*> v;
	v.push_back(root);
	while(num<n){
		if(flag){
			for(int i=0;i<v.size();i++){
				node* now=v[i];
				if(now->lchild!=NULL) q.push(now->lchild);
				if(now->rchild!=NULL) q.push(now->rchild);
			}
			for(int i=v.size()-1;i>=0;i--){
				node* now=v[i];
				zorder[num++]=now->data;
			}
			v.clear();
			flag=false;
		}
		else{
			while(!q.empty()){
				node* now=q.front();
				zorder[num++]=now->data;
				q.pop();
				if(now->lchild!=NULL) v.push_back(now->lchild);
				if(now->rchild!=NULL) v.push_back(now->rchild);
			}
			flag=true;
		}
	}
}
int main(){
  int n;
  scanf("%d",&n);
  for(int i=0;i<n;i++){
  	scanf("%d",&inorder[i]);
  }
  for(int i=0;i<n;i++){
  	scanf("%d",&postorder[i]);
  }
  node* root=create(0,n-1,0,n-1);
  BFS(root,n);
  for(int i=0;i<n;i++){
  	printf("%d",zorder[i]);
  	if(i!=n-1) printf(" ");
  	else printf("\n");
  }
}

这道题首先需要利用中序遍历和后序遍历来构建二叉树，然后再进行层次遍历的变式输出。
一开始我觉得我的构造肯定会出错，因为我这一块总是记得迷迷糊糊的，但是很意外的是这次我的思路异常清晰。
然后关于“z字遍历”我的做法是这样的，用队列来正向输出，然后用了一个vector来模拟栈来反向输出，用一个bool变量来控制输出类型，每次正向或者反向输出之后都将bool取反，具体操作可以看代码，我觉得这个题我的做法应该是比较优秀的。