Implement a binary tree data structure with the following operations:

• new BinaryTree(): Create an instance of a BinaryTree. The root is initialized to null, indicating the tree starts empty.
• size(): Return the number of nodes in the tree. Required time complexity: O(n), where n is the number of nodes in the tree.
• height(): Return the height of the tree, defined as the number of edges on the longest path from the root to a leaf. The height of an empty tree is 0. Required time complexity: O(n).
• inOrder(): Return an array of values from an in-order traversal. Required time complexity: O(n).
• preOrder(): Return an array of values from a pre-order traversal. Required time complexity: O(n).
• postOrder(): Return an array of values from a post-order traversal. Required time complexity: O(n).
• isBalanced(): Return true if the tree is balanced. A binary tree is balanced if, for every node in the tree, the height difference between its left and right subtrees is at most 1. Required time complexity: O(n).
• isComplete(): Return true if the tree is complete. A binary tree is complete if all levels are fully filled, except possibly the last, which must be filled from left to right. Required time complexity: O(n).

Use the helper class BinaryTreeNode to represent nodes, with value, left, and right properties.

Examples

const tree = new BinaryTree();
tree.root = new BinaryTreeNode(10);
tree.root.left = new BinaryTreeNode(5);
tree.root.right = new BinaryTreeNode(15);

tree.size(); // 3
tree.height(); // 1
tree.inOrder(); // [5, 10, 15]
tree.preOrder(); // [10, 5, 15]
tree.postOrder(); // [5, 15, 10]
tree.isBalanced(); // true
tree.isComplete(); // true

Constraints

• 0 <= Number of nodes <= 100
• Each node in a binary tree has at most one left child and one right child