Problem:
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Reconstruction of a Tree

Write a program which reads two sequences of nodes obtained by the preorder tree walk and the inorder tree walk on a binary tree respectively, and prints a sequence of the nodes obtained by the postorder tree walk on the binary tree.

Input

In the first line, an integer $n$, which is the number of nodes in the binary tree, is given.

In the second line, the sequence of node IDs obtained by the preorder tree walk is given separated by space characters.

In the second line, the sequence of node IDs obtained by the inorder tree walk is given separated by space characters.

Every node has a unique ID from $1$ to $n$. Note that the root does not always correspond to $1$.

Output

Print the sequence of node IDs obtained by the postorder tree walk in a line. Put a single space character between adjacent IDs.

Constraints

$1 \leq n \leq 40$

Sample Input 1

5
1 2 3 4 5
3 2 4 1 5

Sample Output 1

3 4 2 5 1

Sample Input 2

4
1 2 3 4
1 2 3 4

Sample Output 2

4 3 2 1

Solution in JAVA (judge verdict: Accepted)

import java.util.*;
import java.io.*;

public class Main {
	static class Node {
		int id = 0;
		Node left = NIL;
		Node right = NIL;
	}
	static Node NIL = new Node();
	
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);

		int n = scanner.nextInt();
		
		Node[] tree = new Node[n + 1];
		tree[0] = NIL;
		NIL.id = 0;
		for (int i = 1; i < n + 1; i ++) {
			tree[i] = new Node();
		}
		int[] preorder = new int[n];
		int[] inorder = new int[n];
		int[] postorder = new int[n];
		
		for (int i = 0; i < n; i ++) {
			preorder[i] = scanner.nextInt();
		}
		for (int i = 0; i < n; i ++) {
			inorder[i] = scanner.nextInt();
		}
		
		scanner.close();
		
		reci = 0;
		reconstruct(preorder, inorder, 0, n - 1, tree);
		
		postorder(tree[preorder[0]], postorder, 0);
		for (int i = 0; i < n - 1; i ++) {
			System.out.print(postorder[i]);
			System.out.print(" ");
		}
		System.out.println(postorder[n - 1]);
	}
	
	static int reci;
	static Node reconstruct(int[] preorder, int[] inorder, int left, int right, Node[] tree) {
		if (left > right) return NIL;
		Node root = tree[preorder[reci]];
		root.id = preorder[reci];
		
		int mid;
		for (mid = left; tree[inorder[mid]] != root; mid ++);
		reci ++;
		
		root.left = reconstruct(preorder, inorder, left, mid - 1, tree);
		root.right = reconstruct(preorder, inorder, mid + 1, right, tree);
		
		return root;
	}
	
	static int postorder(Node u, int[] postorder, int posti) {
		if (u == NIL) return posti;
		
		posti = postorder(u.left, postorder, posti);
		posti = postorder(u.right, postorder, posti);
		postorder[posti] = u.id;
		
		return posti + 1;
	}
}