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

public class Main {
	static class Node {
		int key = 0;
		Node parent = NIL;
		Node left = NIL;
		Node right = NIL;
	}
	static Node NIL = new Node();
	
	static class Tree {
		Node root = NIL;
	}
	
	public static void main(String[]  args) {
		int m = nextInt();
		
		Tree tree = new Tree();
		
		for (int i = 0; i < m; i ++) {
			next();
			switch (nextbuf[0]) {
				case 'i':
					insert(tree, nextInt());
					break;
				case 'f':
					System.out.println(find(tree, nextInt()) ? "yes": "no");
					break;
				default:
					inorder(tree.root);
					System.out.println();
					preorder(tree.root);
					System.out.println();
					break;
			}
		}
	}
	
	static void insert(Tree tree, int key) {
		Node z = new Node();
		z.key = key;
		Node y = NIL;
		Node x = tree.root;
		
		while (x != NIL) {
			y = x;
			x = (key < x.key) ? x.left: x.right;
		}
		z.parent = y;
		
		if (z.parent == NIL) {
			tree.root = z;
		} else if (z.key < z.parent.key) {
			z.parent.left = z;
		} else {
			z.parent.right = z;
		}
	}
	
	static boolean find(Tree tree, int key) {
		Node x = tree.root;

		while (x != NIL) {
			if (key == x.key) return true;
			x = (key < x.key) ? x.left: x.right;
		}
		return false;
	}

	static void preorder(Node u) {
		if (u == NIL) return;

		System.out.print(" ");
		System.out.print(u.key);
		preorder(u.left);
		preorder(u.right);
	}

	static void inorder(Node u) {
		if (u == NIL) return;

		inorder(u.left);
		System.out.print(" ");
		System.out.print(u.key);
		inorder(u.right);
	}
	
	private static byte[] buf = new byte[1 << 20];
	private static int len = 0;
	private static int ptr = 0;

	private static byte read() {
		if (ptr >= len) {
			try {
				len = System.in.read(buf);
			} catch (Exception e) {
			}
			ptr = 0;
		}

		return buf[ptr++];
	}

	private static int nextInt() {
		int num = 0;

		byte b = read();
		while(!Character.isDigit(b) && b != '-') b = read();
		int sign;
		if (b == '-') {
			sign = -1;
			b = read();
			if (!Character.isDigit(b)) return 0;
		} else {
			sign = 1;
		}
		
		while(Character.isDigit(b)) {
			num *= 10;
			num += sign * (b - '0');
			b = read();
		}
		
		return num;
	}

	private static byte[] nextbuf = new byte[11];
	private static int nextlen;

	private static void next() {
		nextlen = 0;

		byte b = read();
		while(!Character.isAlphabetic(b) && !Character.isDigit(b)) b = read();
		while(Character.isAlphabetic(b) || Character.isDigit(b)) {
			nextbuf[nextlen++] = b;
			b = read();
		}
	}
}