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 Ruby (judge verdict: Wrong Answer)

#!/usr/bin/ruby
class TLLI
	def initialize
		@trianies=[nil,[0,0,0]]
	end
	def root_triany
		1
	end
	def allocate_triany
		siz=@trianies.size
		@trianies[siz]=[0,0,0]
		return siz
	end
	def set_a(id,value)
		return if !@trianies[id]
		@trianies[id][0]=value
	end
	def set_b(id,value)
		return if !@trianies[id]
		@trianies[id][1]=value
	end
	def set_c(id,value)
		return if !@trianies[id]
		@trianies[id][2]=value
	end
	def get_a(id)
		return if !@trianies[id]
		@trianies[id][0]
	end
	def get_b(id)
		return if !@trianies[id]
		@trianies[id][1]
	end
	def get_c(id)
		return if !@trianies[id]
		@trianies[id][2]
	end
end

$black={}
class SETI <TLLI # a=>key, b=>left, c=>right
	def set_entry(key)
		x=root_triany
		y=0
		while x!=0&&!(x==root_triany&&get_a(x)==0) #fixme
			y=x
			if key<get_a(x)
				x=get_b(x)
			else
				x=get_c(x)
			end
		end
		if y==0
			set_a(root_triany,key)
		else
			leaf=allocate_triany
			set_a(leaf,key)
			if key<get_a(y)
				set_b(y,leaf)
			else
				set_c(y,leaf)
			end
		end
	end
	def find_entry(key)
		t=root_triany
		loop{
			return true if key==get_a(t)
			if key<get_a(t)
				t=get_b(t)
				return false if t==0
			else
				t=get_c(t)
				return false if t==0
			end
		}
	end
	def delete_entry(key)
		left=false
		prev=nil
		t=root_triany
		loop{
			break if key==get_a(t)
			prev=t
			if key<get_a(t)
				t=get_b(t)
				left=true
				return false if t==0
			else
				t=get_c(t)
				left=false
				return false if t==0
			end
		}
		if get_b(t)>0&&get_c(t)>0
			children=inorder_internal(get_c(t))+[t]
			left=false
			set_a(t,get_a(children[0]))
			prev=children[1]
			return if prev==t
			set_b(prev,0)
			t=children[0]
		end
		if get_b(t)+get_c(t)>0
			n=get_b(t)+get_c(t)
			if left
				set_b(prev,n)
			else
				set_c(prev,n)
			end
		else
			if left
				set_b(prev,0)
			else
				set_c(prev,0)
			end
		end
	end
	def inorder_internal(t)
		t==0 ? [] : inorder_internal(get_b(t))+[t]
	end
	def print_preorder(t)
		return if t==0
		Kernel.print " #{get_a(t)}"
		print_preorder(get_b(t))
		print_preorder(get_c(t))
	end
	def print_inorder(t)
		return if t==0
		print_inorder(get_b(t))
		Kernel.print " #{get_a(t)}"
		print_inorder(get_c(t))
	end
	def print_postorder(t)
		return if t==0
		print_postorder(get_b(t))
		print_postorder(get_c(t))
		Kernel.print " #{get_a(t)}"
	end
	def print
		print_inorder(root_triany);puts
		print_preorder(root_triany);puts
	end
end

seti=SETI.new
$<.drop(1).each{|l|
	a=l.split
	if a[0]=='insert'
		seti.set_entry(a[1].to_i)
	elsif a[0]=='find'
		puts seti.find_entry(a[1].to_i) ? :yes : :no
	elsif a[0]=='delete'
		seti.delete_entry(a[1].to_i)
	elsif a[0]=='print'
		seti.print
	end
}