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 <math.h>
#include <string.h>
#include <stdlib.h>
#define N 200001
#define MAX 1000

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

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

void insert(int k){
    struct Node *y=NI;
    struct Node *x=root;
    struct Node *z;

    z=(struct Node *)malloc(sizeof(struct Node));
    z->key=k;
    z->left=NI;
    z->right=NI;
    while(x != NI){
        y=x;
        if(z->key < x->key){  x=x->left;}
        else{x=x->right;}
    }
    z->parent=y;
    if(y==NI){root=z;}
    else{
        if(z->key < y->key){y->left=z;}
        else{y->right=z;}
    }
}

void inorder(struct Node *u){
    if(u==NI){return;}
    inorder(u->left);
    printf(" %d",u->key);
    inorder(u->right);
}
void preorder(struct Node *u){
    if(u==NI){return;}
    printf(" %d",u->key);
    preorder(u->left);
    preorder(u->right);
}
int main(void){
    int j,i,x;
    char line[7];
    scanf("%d",&j);
    for(i=0;i<j;i++){
        scanf("%s",line);
        if(!strcmp(line,"insert")){
            scanf("%d",&x);
            insert(x);
        }
        else if(!strcmp(line,"print")){
            inorder(root);
            puts("");
            preorder(root);
            puts("");
        }
        else{
            scanf("%d",&x);
            struct Node *temp=find(root,x);
            if(temp != NI){puts("yes");}
            else {puts("no");}
        }
    }
    return 0;
}