A1020 Tree Traversals (25 point(s))

后+中=层

1. 原文

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

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 postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. 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:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample output:

4 1 6 3 5 7 2

2. 解析

后+˙中 = 层

3. AC代码

#include<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
int in[40],post[40];
struct node{
	int index,data;
	node(int x,int y){
		index=x;
		data=y;
	}
};
vector<node> level;
bool cmp(node a,node b){
	return a.index<b.index;
}
void levelOrder(int root,int inl,int inr,int index){
	if (inl>inr)
	{
		return;
	}
	int i = inl;
	for (; i <= inr; ++i){
		if (post[root]==in[i])
		{
			break;
		}
	}
    int right = inr-i;
    level.push_back(node(index,post[root]));
    levelOrder(root-1-right,inl,i-1,2*index+1);
    levelOrder(root-1,i+1,inr,2*index+2);
}
int main(){
	int n;
	scanf("%d",&n);
	for (int i = 0; i < n; ++i)
	{
		scanf("%d",&post[i]);
	}
	for (int i = 0; i < n; ++i)
	{
		scanf("%d",&in[i]);
	}
	levelOrder(n-1,0,n-1,0);
	sort(level.begin(),level.end(),cmp);
	for (int i = 0; i < n; ++i)
	{
		printf("%d",level[i].data);
		if (i<n-1)
		{
			printf(" ");
		}else{
			printf("\n");
		}
	}
	
	return  0;
}