Problem:
Binary Search Tree II

Write a program which performs the following operations to a binary search tree $T$ by adding the find operation to A: Binary Search Tree I.

insert $k$: Insert a node containing $k$ as key into $T$.

find $k$: Report whether $T$ has a node containing $k$.

print: Print the keys of the binary search tree by inorder tree walk and preorder tree walk respectively.

Input

In the first line, the number of operations $m$ is given. In the following $m$ lines, operations represented by insert $k$, find $k$ or print are given.

Output

For each find $k$ operation, print "yes" if $T$ has a node containing $k$, "no" if not.

In addition, for each print operation, print a list of keys obtained by inorder tree walk and preorder tree walk in a line respectively. Put a space character before each key.

Constraints

The number of operations $\leq 500,000$

The number of print operations $\leq 10$.

$-2,000,000,000 \leq key \leq 2,000,000,000$

The height of the binary tree does not exceed 100 if you employ the above pseudo code.

The keys in the binary search tree are all different.

Sample Input 1

10
insert 30
insert 88
insert 12
insert 1
insert 20
find 12
insert 17
insert 25
find 16
print

Sample Output 1

yes
no
1 12 17 20 25 30 88
30 12 1 20 17 25 88

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in JAVA (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		
		Node T = null;
		for(int i = 0; i < n; ++i) {
			String[] str = br.readLine().split(" ");
			if(str[0].equals("insert")){
				T = Node.insert(T, new Node(Integer.parseInt(str[1])));
			}else if(str[0].equals("find")){
				if(Node.find(T, Integer.parseInt(str[1]))){
					System.out.println("yes");
				}else{
					System.out.println("no");
				}
			}else{
				Node.printInOrder(T);
				Node.printPreOrder(T);
			}
		}
	
	}
}

class Node{
	public int key;
	public Node left;
	public Node right;
	public Node p;
	public Node(int key){
		this.key = key;
	}
	public static Node insert(Node T, Node z){
		Node y = null;
		Node x = T;
		while(x != null){
			y = x;
			if(z.key < x.key){
				x = x.left;
			}else{
				x = x.right;
			}
		}
		z.p = y;
		
		if(y == null){
			T = z;
		}else if(z.key < y.key){
			y.left = z;
		}else{
			y.right = z;
		}
		return T;
	}
	public static boolean find(Node T, int key){
		Node x = T;
		while(x != null){
			if(key == x.key){
				return true;
			}else if(key < x.key){
				x = x.left;
			}else{
				x = x.right;
			}
		}
		return false;
	}
	public static void printInOrder(Node T){
		StringBuilder sb = new StringBuilder();
		sb.append(' ');
		inOrder(T, sb);
		System.out.println(sb);
	}
	public static void inOrder(Node z, StringBuilder sb){
		if(z == null){
			return;
		}

		if(z.left != null){
			inOrder(z.left, sb);
			sb.append(' ');
		}
		sb.append(z.key);
		if(z.right != null){
			sb.append(' ');
			inOrder(z.right, sb);
		}
	}
	public static void printPreOrder(Node T){
		StringBuilder sb = new StringBuilder();
		sb.append(' ');
		preOrder(T, sb);
		System.out.println(sb);
	}
	public static void preOrder(Node z, StringBuilder sb){
		if(z == null){
			return;
		}
		sb.append(z.key);
		if(z.left != null){
			sb.append(' ');
			preOrder(z.left, sb);
		}
		if(z.right != null){
			sb.append(' ');
			preOrder(z.right, sb);
		}
	}
}