以下代码能够正确统计二叉树中叶子结点的数量。class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right def count_leaf(root): if not root: return 0 if not root.left and not root.right: return 1 return count_leaf(root.left) + count_leaf(root.right) if __name__ == "__main__": root1 = TreeNode(1) root1.left = TreeNode(2) root1.right = TreeNode(3) root1.left.left = TreeNode(4) root1.left.right = TreeNode(5) root1.right.right = TreeNode(6) print(f"二叉树1的叶子节点数：{count_leaf(root1)}") root2 = TreeNode(1) print(f"二叉树2的叶子节点数：{count_leaf(root2)}") 1 root3 = None print(f"空树的叶子节点数：{count_leaf(root3)}")
A. 正确
B. 错误
正确答案：正确
试卷来源：202512 CCF-GESP编程能力等级认证Python六级真题