PAT 甲级 1127 ZigZagging on a Tree

1127 ZigZagging on a Tree （30 point(s)）

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

经验总结：

emmmm  题意是，根据中序和后序序列，求出这个树的Z形层次遍历，其实只需要在正常的层次遍历中，临时存储每一层的遍历序列，如果根结点是第一层，那么奇数层就将临时存储的遍历序列反过来放进遍历序列中，然后输出最终序列即可~

AC代码

#include <cstdio>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn=40;
int n,m,in[maxn],post[maxn],level[maxn],num=0;
vector<int> temp;
struct node
{
	int data,level;
	node *lchild,*rchild;
};
void BFS(node * root)
{
	queue<node * > q;
	root->level=1;
	int pos=1;
	q.push(root);
	while(q.size())
	{
		node * x=q.front();
		q.pop();
		if(x->level!=pos)
		{
			if(pos%2==1)
				for(int i=temp.size()-1;i>=0;--i)
					level[num++]=temp[i];
			else
				for(int i=0;i<temp.size();++i)
					level[num++]=temp[i];
			temp.clear();
			++pos;
		}
		temp.push_back(x->data);
		if(x->lchild!=NULL)
		{
			x->lchild->level=x->level+1;
			q.push(x->lchild);
		}
		if(x->rchild!=NULL)
		{
			x->rchild->level=x->level+1;
			q.push(x->rchild);
		}
	}
	if(pos%2==1)
		for(int i=temp.size()-1;i>=0;--i)
			level[num++]=temp[i];
	else
		for(int i=0;i<temp.size();++i)
			level[num++]=temp[i];
}
node * create(int inL,int inR,int postL,int postR)
{
	if(inL>inR)
		return NULL;
	int i;
	for(i=inL;i<=inR;++i)
		if(in[i]==post[postR])
			break;
	node * root=new node();
	root->data=post[postR];
	root->lchild=create(inL,i-1,postL,postL+i-1-inL);
	root->rchild=create(i+1,inR,postL+i-inL,postR-1);
	return root;
}
int main()
{
	scanf("%d",&n);
	for(int i=0;i<n;++i)
		scanf("%d",&in[i]);
	for(int i=0;i<n;++i)
		scanf("%d",&post[i]);
	node * root=create(0,n-1,0,n-1);
	BFS(root);
	for(int i=0;i<n;++i)
		printf("%d%c",level[i],i<n-1?' ':'\n');
	return 0;
}