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 <string.h>
#include <stdlib.h>
#define N 500000

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

void insert(int);
void inorder(struct node *);
void preorder(Node *);
int find(int);
void delete(int);

int main(){
  int n,i,j,z;
  char com[10],in[] = "insert",pr[] = "print",fi[] = "find",de[] = "delete";

  scanf("%d",&n);
  root=NULL;
  for(i = 0;i < n;i++){
    scanf("%s",com);
    if(strcmp(com,pr)==0){
      inorder(root);
      printf("\n");
      preorder(root);
      printf("\n");
    }
    else if(strcmp(com,in)==0){
      scanf("%d",&z);
      insert(z);
    }
    else if(strcmp(com,fi)==0){
      scanf("%d",&z);
      if(find(z) == 1) printf("yes\n");
      else printf("no\n");
    }
    else{
      scanf("%d",&z);
      delete(z);
    }
  }
  
  
  return 0;
}

void insert(int key){
  Node *y = NULL;
  Node *x = root;//y->xの親 x->root
  Node *z;

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

  if(y == NULL) root = z;
  else if(z->key < y->key) y->left = z;
  else y->right = z;
}

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

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

int find(int z){
  Node *x = root;
  while(1){
  if(z == x->key) return 1;
  else if(z > x->key) x = x->right;
  else x = x->left;
  if(x == NULL) return 0;
  }
}

void delete(int delkey){
  Node *y;
  Node *x = root;
  Node *z;
  Node *child;
  
  
   while(1){
  if(delkey == x->key) break;
  else if(delkey > x->key) x = x->right;
  else x = x->left;
  }
   z = x;
  if(z->left == NULL || z->right == NULL){
    y = z;
    if(z->left != NULL) child = z->left;
    else child = z->right;
  }
  else{
    x = x->right;
    while(1){
      if(x->left != NULL) x = x->left;
      else break;
    }
    y = x;
  }
  if(y == z){
    if(child != NULL){
      child->parent = y->parent;
    }
    if(y->key < y->parent->key){
      y->parent->left = child;
    }
    else y->parent->right = child;
  }
  else{
    z->key = y->key;
    if(y->key < y->parent->key){
      y->parent->left = NULL;
    }
    else y->parent->right = NULL;
  }
  free(y);
}