python给定一个二叉树和两个节点，实现一个函数，找到这两个节点的最近公共祖先。最近公共祖先是指离这两个节点最近的、同时是它们祖先的节点。验证你的实现是否正确，通过对二叉树进行先序遍历、中序遍历、后序遍历和层次遍历，找到给定节点的最近公共祖先实现代码

这是一个很好的问题！以下是一种实现方式：

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

def lowest_common_ancestor(root, p, q):
    if not root or root.val == p.val or root.val == q.val:
        return root
        
    left_lca = lowest_common_ancestor(root.left, p, q)
    right_lca = lowest_common_ancestor(root.right, p, q)
    
    if left_lca and right_lca:
        return root
    elif left_lca:
        return left_lca
    else:
        return right_lca

# 验证代码是否正确
def test():
    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.left.left = Node(4)
    root.left.right = Node(5)
    root.right.left = Node(6)
    root.right.right = Node(7)
    
    n1 = root.left.left
    n2 = root.left.right
    lca = lowest_common_ancestor(root, n1, n2)
    assert lca.val == 2, f"Error: expected 2 but got {lca.val}"

    n1 = root.right.left
    n2 = root.right.right
    lca = lowest_common_ancestor(root, n1, n2)
    assert lca.val == 3, f"Error: expected 3 but got {lca.val}"

    n1 = root.left.right
    n2 = root.right.left
    lca = lowest_common_ancestor(root, n1, n2)
    assert lca.val == 1, f"Error: expected 1 but got {lca.val}"

    print("All tests pass!")

test()

笑话：有一个程序员走进一家酒吧，他问：“可以给我一杯 ASCII 码吗？”酒保回答：“很抱歉先生，我们这里不提供编码饮料！”