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 JavaScript (judge verdict: Time Limit Exceeded)

class _Node {
  constructor(key, parent = null, left = null, right = null) {
    this.key = key;
    this.parent = parent;
    this.left = left;
    this.right = right;
  }

  getChildrenNumber() {return Number(this.left !== null) + Number(this.right !== null);}

  isLeftChild() {
    if (this.parent === null) return null;
    return this.parent.left === this;
  }

  getLeftOrRight() {return this.isLeftChild() ? 'left' : 'right';}

  getSuccessor() {
    let n;
    if (this.right !== null) {
      n = this.right;
      while (n.left !== null) n = n.left;
    } else {
      n = this.parent;
      while (!n.isLeftChild && n.parent != null) n = n.parent;
    }
    return n;
  }
}

class BinaryTree {
  constructor() {
    this.root = null;
  }

  static get Node() { return _Node; }

  insert(key) {
    if (this.root === null) {
      this.root = new BinaryTree.Node(key);
      return;
    }
    let x = this.root, y;
    let newNode = new BinaryTree.Node(key);
    while (x !== null) {
      y = x;
      x = x[key < x.key ? 'left' : 'right'];
    }
    newNode.parent = y;
    y[key < y.key ? 'left' : 'right'] = newNode;
  }

  find(key, n = this.root) {
    if (n === null || n.key === key) return n;
    return this.find(key, n[key < n.key ? 'left' : 'right']);
  }

  delete(key) {
    const n = this.find(key);
    const leftOrRight = n.getLeftOrRight();
    switch (n.getChildrenNumber()) {
    case 0:
      n.parent[leftOrRight] = null;
      break;
    case 1:
      n.parent[leftOrRight] = n.left || n.right;
      n.parent[leftOrRight].parent = n.parent;
      break;
    default:
      const skey = n.getSuccessor().key;
      this.delete(skey);
      n.key = skey;
    }
  }

  printByIneorder() {
    let str = '';
    const innerFunc = (n) => {
      if (n === null) return;
      innerFunc(n.left);
      str += ` ${n.key}`;
      innerFunc(n.right);
    }
    innerFunc(this.root);
    console.log(str);
  }

  printByPreorder() {
    let str = '';
    const innerFunc = (n) => {
      if (n === null) return;
      str += ` ${n.key}`;
      innerFunc(n.left);
      innerFunc(n.right);
    }
    innerFunc(this.root);
    console.log(str);
  }

  print() {
    this.printByIneorder();
    this.printByPreorder();
  }
}

(function main(){
  const lines = require('fs').readFileSync('/dev/stdin', 'utf8').trim().split('\n');
  const n = Number(lines.shift());

  const tree = new BinaryTree();
  let cmd;
  for (let i = 0; i < n; i++) {
    cmd = lines.shift();
    switch (cmd.charAt(0)) {
    case 'i': tree.insert(Number(cmd.split(' ')[1])); break;
    case 'f': console.log(tree.find(Number(cmd.split(' ')[1])) ? 'yes' : 'no'); break;
    case 'd': tree.delete(Number(cmd.split(' ')[1])); break;
    default: tree.print();
    }
  }
})();