1127. ZigZagging on a Tree (30)

​ 时间限制

​ 400 ms

​ 内存限制

​ 65536 kB

​ 代码长度限制

​ 16000 B

​ 判题程序

​ Standard

​ 作者

​ CHEN, Yue

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

此题为中序后序求层次遍历，

然后以s形状输出层次遍历

利用result存层次，depth存每一层次有多少点

然后根据depth输出

代码如下：

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


struct Node{
	int key;
	int index;
}node;
int cmp1(Node a,Node b){
	return a.index<b.index;
}
vector<int>in,post;
vector<int>depth(50);
vector<Node>result;
void print_result(int root,int start,int end,int index,int deep){
	if(start>end){
		return;
	}
	int i=start;
	for(;i<end;++i){
		if(in[i]==post[root]){
			break;
		}
	}
	node.index=index;node.key=post[root];
	result.push_back(node);
	depth[deep]++;
	
	print_result(root-(end-i+1),start,i-1,2*index+1,deep+1);
	print_result(root-1,i+1,end,2*index+2,deep+1);
}

int main(){
	int n;
	scanf("%d",&n);
	post.resize(n);
	in.resize(n);
	for(int i=0;i<n;++i){
		scanf("%d",&in[i]);
	}
	for(int i=0;i<n;++i){
		scanf("%d",&post[i]);
	}
	print_result(n-1,0,n-1,0,0);
	
	sort(result.begin(),result.end(),cmp1);
	
	int index=1;
	printf("%d",result[0].key);
	for(int i=1;i<n;++i){
		if(depth[i]>0){
			int q=depth[i];
			if(i%2!=0){
				for(int j=index;j<q+index;++j){
					printf(" %d",result[j].key);
				}
				index+=q;
			}else{
				for(int j=index+q-1;j>=index;--j){
					printf(" %d",result[j].key);
				}
				index+=q;
			}
		}else{
			break;
		}
	}
	return 0;
}