在给出的程序框架里填写4个二叉树遍历应用算法函数，包括输出二叉树中度为2的结点值，求二叉树的高度，求二叉树中结点个数，求二叉树中叶子结点个数。 （后两个使用全局变量n和n0分 别进行统计个数）

class Node:
    def __init__(self, value=None, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

class BinaryTree:
    def __init__(self, root=None):
        self.root = root

    # 输出二叉树中度为2的结点值
    def print_full_node(self):
        def dfs(node):
            if not node:
                return
            if node.left and node.right:
                print(node.value)
            dfs(node.left)
            dfs(node.right)
        dfs(self.root)

    # 求二叉树的高度
    def get_height(self):
        def dfs(node):
            if not node:
                return 0
            left_height = dfs(node.left)
            right_height = dfs(node.right)
            return 1 + max(left_height, right_height)
        return dfs(self.root)

    # 求二叉树中结点个数
    def get_node_count(self):
        def dfs(node):
            if not node:
                return 0
            return 1 + dfs(node.left) + dfs(node.right)
        return dfs(self.root)

    # 求二叉树中叶子结点个数（使用全局变量n0）
    def get_leaf_count(self):
        global n0
        n0 = 0
        def dfs(node):
            global n0
            if not node:
                return
            if not node.left and not node.right:
                n0 += 1
            dfs(node.left)
            dfs(node.right)
        dfs(self.root)
        return n0

if __name__ == '__main__':
    # 构造一棵二叉树
    n1 = Node('A')
    n2 = Node('B')
    n3 = Node('C')
    n4 = Node('D')
    n5 = Node('E')
    n6 = Node('F')
    n7 = Node('G')
    n8 = Node('H')
    n1.left, n1.right = n2, n3
    n2.left, n2.right = n4, n5
    n3.left, n3.right = n6, n7
    n5.left = n8
    tree = BinaryTree(n1)

    # 输出二叉树中度为2的结点值
    tree.print_full_node()

    # 求二叉树的高度
    print(tree.get_height())

    # 求二叉树中结点个数
    print(tree.get_node_count())

    # 求二叉树中叶子结点个数
    print(tree.get_leaf_count())