Problem:
Binary Search Tree III

Write a program which performs the following operations to a binary search tree $T$ by adding delete operation to B: Binary Search Tree II.

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

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

delete $k$: Delete a node containing $k$.

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

The operation delete $k$ for deleting a given node $z$ containing key $k$ from $T$ can be implemented by an algorithm which considers the following cases:

If $z$ has no children, we modify its parent $z.p$ to replace $z$ with NIL as its child (delete $z$).

If $z$ has only a single child, we "splice out" $z$ by making a new link between its child and its parent.

If $z$ has two children, we splice out $z$'s successor $y$ and replace $z$'s key with $y$'s key.

Input

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

18
insert 8
insert 2
insert 3
insert 7
insert 22
insert 1
find 1
find 2
find 3
find 4
find 5
find 6
find 7
find 8
print
delete 3
delete 7
print

Sample Output 1

yes
yes
yes
no
no
no
yes
yes
1 2 3 7 8 22
8 2 1 3 7 22
1 2 8 22
8 2 1 22

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>
 struct node{

  struct node *right,*left,*parent;
  int key;
};
typedef struct node * Node; 
#define NIL NULL
Node root;

void insert(int);
void inorder(Node);
void preorder(Node);
Node find( Node,int);
Node treemin(Node);
Node successor(Node);
void treedel(Node);

int main(){
  int n,x,i;
  char str[20];
  Node get;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%s",str);
    if(str[0]=='f') {
      scanf("%d",&x);
      get = find(root,x);
      if(get != NIL) printf("yes\n");
      else printf("no\n");
    }else if(strcmp(str,"insert")==0){
      scanf("%d",&x);
      insert(x);
    }else if(strcmp(str,"print")==0 ){
      inorder(root);
      printf("\n");
      preorder(root);
      printf("\n");
    }else if(strcmp(str,"delete")==0){
      scanf("%d",&x);
      treedel(find(root,x));
    }
  }

  return 0;
  }


void insert(int k){
  Node y,x,z;

  y= NIL; //x's parent
  x=root;
  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 x){
  if(x == NIL) return;
  inorder(x->left);
  printf(" %d",x->key);
  inorder(x->right);
}
void preorder(Node x){
  if(x == NIL) return;
  printf(" %d",x->key);
  preorder(x->left);
  preorder(x->right);


}

Node find(Node x,int k){

  while(x != NIL&&k!=x->key){
    if(k < x->key) x = x->left;
    else x = x->right;

  }


  return x;
}

Node treemin(Node x){
  while(x->left !=NIL) x =x->left;
  
  return x;
}

Node successor(Node x){
  Node y;

  if(x->right != NIL) return treemin(x->right);
  y=x->parent;
  while(y!=NIL&&x==y->right){
    x=y;
    y=y->parent;
  }
  return y;
}

void treedel(Node k){
  Node x,y;

  if(k->left == NIL||k->right == NIL ) y = k;
  else y=successor(k);

  if(y->left != NIL) x=y->left;
  else x=y->right;

  if(x != NIL){
    x->parent = y->parent;
  }

  if(y->parent == NIL){
    root = x;
  }else{
    if(y==y->parent->left){
      y->parent->left=x;
    }else{
      y->parent->right=x;
    }
  }
  if(y != k) k->key = y->key;

  free(y);
}