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>
#define NIL NULL

struct node{
  struct node *right;
  struct node *left;
  struct node *parent;
  int key;
};
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 = malloc(sizeof(struct node));
  z->key = k;
  z->left = NIL;
  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 i,n,x;
    char com[20];
    Node t;

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