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)

// AOJ ALDS1_8_C Binary Search Tree III
// 2018.5.1 bal4u

#include <stdio.h>

#if 1
#define gc() getchar_unlocked()
#define pc(c) putchar_unlocked(c)
#else
#define gc() getchar()
#define pc(c) putchar(c)
#endif

int in()
{
	int n = 0, c = gc();
	if (c == '-') { c = gc();
		do n = 10*n + (c & 0xf), c = gc(); while (c >= '0');
		return -n;
	}
	do n = 10*n + (c & 0xf), c = gc(); while (c >= '0');
	return n;
}

void skip(int k) { while (k--) gc(); }

void out(int n)
{
	int i;
	char ob[20];

	if (!n) { pc('0'); return; }
	if (n < 0) pc('-'), n = -n;
	i = 0; while (n) ob[i++] = n%10 + '0', n/=10;
	while (i--) pc(ob[i]); 
}

void sout(char *s) { while (*s) pc(*s++); }

typedef struct { int key; int parent; int left, right; } BT;
BT bt[500005]; int sz;
int root;

void preorder(int node)
{
	pc(' '), out(bt[node].key);
	if (bt[node].left  > 0) preorder(bt[node].left);
	if (bt[node].right > 0) preorder(bt[node].right);
}

void inorder(int node)
{
	if (bt[node].left  > 0) inorder(bt[node].left);
	pc(' '), out(bt[node].key);
	if (bt[node].right > 0) inorder(bt[node].right);
}

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

	y = 0, x = root;
	while (x) {
		y = x;
		if (bt[z].key < bt[x].key) x = bt[x].left;
		else                       x = bt[x].right;
	}
	bt[z].parent = y;

	if (!y) root = z;
	else if (bt[z].key < bt[y].key) bt[y].left  = z;
	else                            bt[y].right = z;
}

int find(int node, int key)
{
	int t;

	t = key - bt[node].key;
	if (!t) return node;
	if (t < 0) {
		if (bt[node].left > 0) return find(bt[node].left, key);
	} else {
		if (bt[node].right > 0) return find(bt[node].right, key);
	}
	return -1;
}

void delete(int z)
{
	int p, c, y;

	if (bt[z].left <= 0 && bt[z].right <= 0) {
		p = bt[z].parent;
		if (z == bt[p].left) bt[p].left  = 0;
		else                 bt[p].right = 0;
	} else if (bt[z].left > 0 && bt[z].right > 0) {
		y = bt[z].right;
		while (bt[y].left > 0) y = bt[y].left;
		bt[z].key = bt[y].key;
		delete(y);
	} else {
		p = bt[z].parent;
		if (bt[z].left > 0) c = bt[z].left;
		else                c = bt[z].right;
		if (z == bt[p].left) bt[p].left  = c;
		else                 bt[p].right = c;
		bt[c].parent = p;
	}
}

int main()
{
	int n, i;
	char op;

	n = in(), root = 0, sz = 1;
	for (i = 0; i < n; i++) {
		op = gc();
		if (op == 'p') {            // print
			skip(5);
			inorder(root), pc('\n');
			preorder(root), pc('\n');
		} else if (op == 'i') {     // insert
			skip(6);
			bt[sz].key = in();
			insert(sz++);
		} else if (op == 'f') {     // find
			skip(4);
			sout(find(root, in()) >= 0? "yes": "no"), pc('\n');
		} else {                    // delete
			int z;
			skip(6);
			z = find(root, in());
			if (z >= 0) delete(z);
		}
	}
	return 0;
}