【PAT】1127. ZigZagging on a Tree (30)【树的遍历】

题目描述

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.

翻译：假设二叉树中的所有键都是不同的正整数。一棵唯一的二叉树可以由给定的一对后序遍历和先序遍历序列确定。然后按层次顺序打印数字，这是一个简单的标准例程。然而，如果你认为问题太简单，那你就太天真了。这一次，您应该按照“之字形顺序”打印数字——也就是说，从根开始，逐层打印数字，在左到右和右到左之间交替。例如，对于下面的树，您必须输出: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.

翻译：每个输入文件包含一组测试数据。对于每组测试数据，第一行给出一个正整数N(≤30),代表二叉树中的总结点数。第二行给出中序遍历序列，第三行给出后序遍历数列。一行内所有数字之间用空格隔开。

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.

翻译：对于每个测试文件，输出一行该树的之字形数列。一行内所有数字之间用空格隔开，并且行尾不能有多余空格。

---

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

---

解题思路

通过后序遍历和中序遍历弄出树，然后将每一层的节点保存到数组里，再根据第几行选择是正序还是倒序输出。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<vector>
#include<algorithm>
#define INF 99999999
using namespace std;
int N,In[35],Post[35],Hash[35],Step[35]; 
int ccount=0,MaxStep=0;
vector<int> v[31];
void dfs(int a,int b,int step){
	if(a==b){
		Step[a]=step+1;
		MaxStep=max(MaxStep,step+1);
		ccount--;
		return ;
	}
	int tmp=Hash[Post[ccount]];
		if(ccount>=0){
			Step[tmp]=step+1;
			ccount--;
			if(tmp<b) dfs(tmp+1,b,step+1);
			if(tmp>a) dfs(a,tmp-1,step+1);
		}
}
int main(){
	scanf("%d",&N);
	for(int i=0;i<N;i++){
		scanf("%d",&In[i]);Hash[In[i]]=i;
	} 
	for(int i=0;i<N;i++){
		scanf("%d",&Post[i]);
	}
	ccount=N-1;
	dfs(0,N-1,0);
	for(int i=0;i<N;i++){
		v[Step[i]].push_back(In[i]);
	}
	printf("%d",v[1][0]);
	for(int i=2;i<=MaxStep;i++){
		if(i%2==1)
		for(int j=v[i].size()-1;j>=0;j--)printf(" %d",v[i][j]);
		else
		for(int j=0;j<v[i].size();j++)printf(" %d",v[i][j]);		
	}
	printf("\n");
	exit(0);
}