PAT甲级 -- 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

我的思路：

1. 建树，加一个变量level，开一个vector<int> v[1000], v[level]表示第level层的元素，一开始level为0，每一次level+1。

2. 之后之字形输出 

#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;

int n;
vector<int> inorder, postorder;
vector<int> v[10000];

void preBuild(int inl, int inr, int postRoot, int level){
	if(inl > inr)  return；
	int i = inl;
	while(inl < inr && postorder[postRoot] != inorder[i]) i++;
	v[level].push_back(inorder[i]);
	preBuild(inl, i-1, postRoot-1+i-inr,level+1);
	preBuild(i+1,inr, postRoot-1, level+1);
}

int main(){
	scanf("%d", &n);
	inorder.resize(n), postorder.resize(n);
	for(int i = 0; i < n; i++) scanf("%d", &inorder[i]);
	for(int i = 0; i < n; i++) scanf("%d", &postorder[i]);
	preBuild(0,n-1,n-1,0);
	printf("%d", v[0][0]);
	for (int i = 1; i < 35; i++){
		if (i % 2 == 1){ 
			for(int j = 0; j < v[i].size(); j++){
				printf(" %d",v[i][j]);
				
			}
						
		}else{
			for (int j = v[i].size()-1; j >= 0; j--){
				printf(" %d",v[i][j]);
			}
			
		}

		
	}
	
	return 0;
}