PAT甲级1127 ZigZagging on a Tree (30)

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

根据中序和后序建树，输出层序遍历，但是要按单数层行从右往左， 双数层行从左往右输出。

用一个vector二维数组保存每层的信息，遇到单数层从后往前输出，遇到双数层从前往后输出

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

struct node{
	int data,level;
	node* left,*right;
	node(int val){
		data=val;
		right=NULL;
		left=NULL;
	}
};

const int maxn=35;
int post[maxn],in[maxn];
int n;
vector<int> res[maxn];

node* buildtree(node* r,int inl,int inr,int postl,int postr){
	if(inl>inr) return NULL;
	int root=post[postr];
	if(r==NULL) r=new node(root);
	int k=0;
	while(root!=in[k]) k++;
	int leftnum=k-inl;
	r->left =buildtree( r->left,inl,k-1,postl,postl+leftnum-1);
	r->right=buildtree(r->right,k+1,inr,postl+leftnum,postr-1);
	return r;
}

int maxlevel=0;
void dfs(node* root,int l){
	if(root==NULL) return;
	if(l>maxlevel) maxlevel=l;
	root->level=l;
	res[root->level].push_back(root->data);
	dfs(root->left,l+1);
	dfs(root->right,l+1);
}

int main(){
	cin>>n;
	for(int i=0;i<n;i++){
		cin>>in[i];
	}
	for(int i=0;i<n;i++){
		cin>>post[i];
	}
	node* root=NULL;
	root=buildtree(root,0,n-1,0,n-1);
	dfs(root,1);
	for(int i=1;i<=maxlevel;i++){
		if(i%2==0){
			for(int j=0;j<res[i].size();j++){
				printf(" %d",res[i][j]);
			}
		}else{
			for(int j=res[i].size()-1;j>=0;j--){
				if(i==1&&j==0) printf("%d",res[i][j]);
				else printf(" %d",res[i][j]);
			}
		}
	}

	return 0;
}