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 C (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
//#include <iostream>
//using namespace std;

#define NIL NULL

struct node{
  int key;
  struct node *right, *left, *parent;
};

typedef struct node * Node;

Node root;

Node find(Node u, int k){
  while(u!=NIL&&k!=u->key){
    if(k<u->key) u=u->left;
    else u=u->right;
  }
  return u;
}

void insert(int k){
  Node y = NIL;
  Node x = root;
  Node z;
  z = /*(Node *)*/malloc(sizeof(struct node));
  z->key = k;
  z->right = NIL;
  while(x!=NIL){
    y = x;
    if(z->key<x->key) x=x->left;
    else x=x->right;
  }
  z->parent = y;
  if(y==NIL) root=z;
  else{
    if(z->key<y->key) y->left=z;
    else y->right=z;
  }
}

void inorder(Node u){
  if(u==NIL) return;
  inorder(u->left);
  printf(" %d",u->key);
  inorder(u->right);
}

void preorder(Node u){
  if(u==NIL) return;
  printf(" %d",u->key);
  preorder(u->left);
  preorder(u->right);
}

int main(){
  int n, i, x;
  char com[10];

  scanf("%d", &n);

  for(i=0; i<n; i++){
    scanf("%s", com);
    if(com[0]=='f'){
      scanf("%d", &x);
      Node t = find(root, x);
      if(t!=NIL) printf("yes\n");
      else printf("no\n");
    }
    else if(!strcmp(com, "insert")){
      scanf("%d", &x);
      insert(x);
    }
    else if(!strcmp(com, "print")){
      inorder(root);
      printf("\n");
      preorder(root);
      printf("\n");
    }
  }

  return 0;
}