如何获取二叉树中节点的最小祖先

# 定义二叉树节点
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def findMinAncestor(root, p, q):
    # 存储节点和父节点对应关系的字典
    parent = {}

    # DFS遍历二叉树，记录父节点信息
    def dfs(node, par):
        if node:
            parent[node.val] = par
            dfs(node.left, node)
            dfs(node.right, node)

    # 开始DFS遍历
    dfs(root, None)

    # 获取节点p的所有祖先节点
    ancestors = set()
    while p:
        ancestors.add(p.val)
        p = parent[p.val]

    # 从节点q开始向上遍历，找到最小祖先节点
    while q:
        if q.val in ancestors:
            return q
        q = parent[q.val]

    return None

# 创建二叉树
root = TreeNode(3)
root.left = TreeNode(5)
root.right = TreeNode(1)
root.left.left = TreeNode(6)
root.left.right = TreeNode(2)
root.right.left = TreeNode(0)
root.right.right = TreeNode(8)
root.left.right.left = TreeNode(7)
root.left.right.right = TreeNode(4)

# 节点p和节点q的值
p_val = 5
q_val = 4

# 查找最小祖先节点
result = findMinAncestor(root, TreeNode(p_val), TreeNode(q_val))

if result:
    print("节点 {} 和节点 {} 的最小祖先节点值为: {}".format(p_val, q_val, result.val))
else:
    print("未找到最小祖先节点")