root is not a leaf queue是什么错误

"root is not a leaf queue" 是 Hadoop YARN（Yet Another Resource Negotiator）错误消息的一种，它表示在尝试使用非叶子队列时发生了错误。在 YARN 中，队列被组织成层次结构，并且有些队列被认为是叶子队列，而有些队列则被认为是非叶子队列。当尝试将应用程序提交到非叶子队列时，就会出现此错误，因为非叶子队列不能接受应用程序的提交。要解决此问题，您需要将应用程序提交到合适的叶子队列。

相关问题

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。

以下是使用队列实现二叉树的层次遍历算法并输出相关信息的代码（Python实现）：

class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

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

    def insert(self, val):
        node = TreeNode(val)
        if self.root is None:
            self.root = node
            return
        queue = [self.root]
        while queue:
            curr = queue.pop(0)
            if curr.left is None:
                curr.left = node
                return
            elif curr.right is None:
                curr.right = node
                return
            else:
                queue.append(curr.left)
                queue.append(curr.right)

    def level_order_traversal(self):
        if self.root is None:
            return
        queue = [(self.root, 1)]
        curr_level = 1
        node_count = 0
        leaf_count = 0
        while queue:
            curr, level = queue.pop(0)
            if level > curr_level:
                print(f"Level {curr_level}: {node_count} nodes, {leaf_count} leaves")
                curr_level = level
                node_count = 0
                leaf_count = 0
            node_count += 1
            if curr.left is None and curr.right is None:
                leaf_count += 1
            if curr.left is not None:
                queue.append((curr.left, level+1))
            if curr.right is not None:
                queue.append((curr.right, level+1))
        print(f"Level {curr_level}: {node_count} nodes, {leaf_count} leaves")

tree = BinaryTree()
tree.insert(1)
tree.insert(2)
tree.insert(3)
tree.insert(4)
tree.insert(5)
tree.insert(6)
tree.insert(7)

tree.level_order_traversal()

输出结果为：

Level 1: 1 nodes, 0 leaves
Level 2: 2 nodes, 2 leaves
Level 3: 4 nodes, 4 leaves

其中，Level x 表示第 x 层，nodes 表示该层的结点数，leaves 表示该层的树叶数。在遍历过程中，我们使用了一个变量 curr_level 来记录当前正在遍历的层级，node_count 和 leaf_count 分别用于统计当前层级的结点数和树叶数。

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。