A22254. 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.l…
判断题
困难
知识点
题目描述
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)}")选项(单选)
答案解析
详细答案解析为会员权益,按每日次数查看。
开通 / 升级会员
上一题
下一题