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>

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

Node root;

void insert(int);
void inorder(Node);
void preorder(Node);
Node treeSearch(Node,int);
Node treeMinimum(Node);
Node treeSuccessor(Node);
void treeDelete(Node);

int main(){
  int n, i, x;
  char com[20];
  scanf("%d",&n);

  for (i=0;i<n;i++){
    scanf("%s",com);
    if(com[0]=='f'){
      scanf("%d",&x);
      Node t = treeSearch(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");
    }

    else if(com[0]=='d'){
      scanf("%d",&x);
      treeDelete(treeSearch(root,x));
    }
  }

  return 0;
}


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){
    inorder(u->left);
    printf(" %d",u->key);
    inorder(u->right);
  }

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

Node treeSearch(Node u, int k){
  if(u==NIL || k==u->key) return u;
    
  if(k<u->key) return treeSearch(u->left,k);
  else return treeSearch(u->right,k);
  
}

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

Node treeSuccessor(Node x){
  Node y;
  
  if(x->right!=NIL) return treeMinimum(x->right);

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

void treeDelete(Node z){
  Node y;
  Node x;

  if(z->left==NIL || z->right==NIL) y=z;
  else y=treeSuccessor(z);

  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!=z) z->key=y->key;

  free(y);
  
}