1127 ZigZagging on a Tree (30分)

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

题目描述：已知中序遍历和后序遍历可以得到唯一的二叉树；已知中序遍历和先序遍历，可以得到唯一的二叉树。现在题目给出中序和后序遍历，需要求出层序遍历的结果。但是这个层序遍历需要按照Z字行输出。

解题思路：不需要建树，递归就能得到层序遍历的结果。1086题解解释了已知先序遍历和中序遍历，求后序遍历的结果，思路与这题差不多，可以参考。

#include<bits/stdc++.h>
using namespace std;
int n, cnt;
const int maxn = 1e4+5;
vector<int> inorder, post;
vector<int> level[maxn];
void get_level(int root, int inl, int inr, int index){
	if(inl>inr) return ;
	int i = inl;
	cnt = index;
	level[index].push_back(post[root]);
	while(i<=inr&&inorder[i]!=post[root]) i++;
	get_level(root-(inr-i+1),inl,i-1,index+1);
	get_level(root-1,i+1,inr,index+1);
}
int main()
{
	scanf("%d", &n);
	inorder.resize(n);
	post.resize(n);
	for(int i=0; i<n; i++) scanf("%d", &inorder[i]);
	for(int i=0; i<n; i++) scanf("%d", &post[i]);
	get_level(n-1,0,n-1,1);
	for(int i=1; i<=cnt; i++){
		if(i%2==0){
			for(int j=0; j<level[i].size(); j++){
				if(i!=1) printf(" ");
				printf("%d", level[i][j]);
			}
		}
		else{
			for(int j=level[i].size()-1; j>=0; j--){
				if(i!=1) printf(" ");
				printf("%d", level[i][j]);
			}
		}
	}
	return 0;
}