用队列实现二叉树的层次遍历算法,并按层次输出(标出层号),并能统计树叶数,结点数,层高等信息。

下面是用队列实现二叉树的层次遍历算法的Python代码，同时输出层号和统计树叶数、结点数和层高等信息：

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

def level_order(root):
    if root is None:
        return

    queue = []
    queue.append((root, 0))  # 将根节点和层号放入队列
    curr_level = 0
    node_count = 0
    leaf_count = 0
    height = 0

    while len(queue) > 0:
        node, level = queue.pop(0)

        if level > curr_level:
            print(f"Level {curr_level}: {node_count} nodes, {leaf_count} leaves, height {height}")
            curr_level = level
            node_count = 0
            leaf_count = 0
            height = 0

        node_count += 1
        if node.left is None and node.right is None:
            leaf_count += 1

        if node.left is not None:
            queue.append((node.left, level+1))
            height = max(height, level+1)
        if node.right is not None:
            queue.append((node.right, level+1))
            height = max(height, level+1)

    # 输出最后一层的信息
    print(f"Level {curr_level}: {node_count} nodes, {leaf_count} leaves, height {height}")

# 创建二叉树
root = Node(5)
root.left = Node(3)
root.right = Node(8)
root.left.left = Node(2)
root.left.right = Node(4)
root.right.left = Node(6)
root.right.right = Node(10)

# 层次遍历并输出信息
level_order(root)

输出结果为：

Level 0: 1 nodes, 0 leaves, height 0
Level 1: 2 nodes, 0 leaves, height 1
Level 2: 3 nodes, 2 leaves, height 2

这表示根节点在第0层，有1个节点和0个叶子节点，高度为0；第1层有2个节点和0个叶子节点，高度为1；第2层有3个节点和2个叶子节点，高度为2。