PAT 1020. Tree Traversals

PAT 1020. Tree Traversals

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

分析

首先后序遍历的最后一个数必定是根节点，然后在中序遍历中找的根节点，根节点前面的是该节点的左子树，后面的是右子树，再对左右子树递归的调用前面的步骤；

代码如下

#include<iostream>
#include<vector>
using namespace std;
vector<int> post,in,level(100000,-1);
void pre(int root,int s,int e,int index){
	if(s>e) return ;
	level[index]=post[root];
	int i=s;
	for(;i<=e;i++)
	if(in[i]==post[root]) break;
	pre(root-(e-i+1),s,i-1,index*2+1);
	pre(root-1,i+1,e,index*2+2);
}
int main(){
	int N; cin>>N;
	post.resize(N); in.resize(N);
	for(int i=0;i<N;i++) cin>>post[i];
	for(int i=0;i<N;i++) cin>>in[i];
	pre(N-1,0,N-1,0);
	for(int i=0;i<level.size();i++)
	if(level[i]!=-1)
	i>0?cout<<" "<<level[i]:cout<<level[i];
	return 0;
}