1127 ZigZagging on a Tree——PAT甲级

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

 solution:
 

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
typedef long long ll;
typedef struct node *tree;
struct node
{
	int val;
	tree left,right;
};
vector<int>ans;
const int maxn=1e5+5;
int n;
int in[maxn],post[maxn];
vector<vector<int> >ret;
void input()
{
	cin>>n;
	for(int i=0;i<n;i++)cin>>in[i];
	for(int i=0;i<n;i++)cin>>post[i];
}
tree build(int in[],int post[],int n)
{
	if(!n)return NULL;
	tree root=new node;
	root->val=post[n-1];
	root->left=root->right=NULL;
	int i;
	for(i=0;i<n;i++)
	{
		if(in[i]==post[n-1])break;
	}
	root->left=build(in,post,i);
	root->right=build(in+i+1,post+i,n-1-i);
	return root;
}
void layerorder(tree root)
{
	if(root==NULL)return;
	queue<tree>q;
	q.push(root);
	while(!q.empty())
	{
		int cnt=q.size();
		vector<int>tmp;
		for(int i=0;i<cnt;i++)
		{
			tree u=q.front();q.pop();
			tmp.push_back(u->val);
			if(u->left)q.push(u->left);
			if(u->right)q.push(u->right);
		}
		ret.push_back(tmp);
	}
}
void output()
{
	for(int i=0;i<ret.size();i++)
	{
		/*
		for(int j=0;j<ret[i].size();j++)
		{
			cout<<ret[i][j]<<' ';
		}
		cout<<endl;
		*/
		
		if(i%2==0)
		{
			for(int j=ret[i].size()-1;j>=0;j--)
			{
				ans.push_back(ret[i][j]);
			}
		}
		
		else
		{
			for(int j=0;j<ret[i].size();j++)
			{
				ans.push_back(ret[i][j]);
			}
		}
		
	}
	
	for(int i=0;i<ans.size();i++)
	{
		if(i)cout<<' ';
		cout<<ans[i];
	}
	
}
int main()
{
	input();
	tree root=build(in,post,n);
	layerorder(root);
	output();
}

         总结：根据中序后序建树，然后层序遍历