Lowest common ancestor of a binary tree(Python)

Problem:
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.
Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note:

All of the nodes’ values will be unique.
p and q are different and both values will exist in the binary tree.

Resource：LeetCode
Link：https://leetcode-cn.com/problems/lowest-common-ancestor-of-a-binary-tree

Analyse:
First,I desided to use the method of recursion.
Second,I cross the river by feeling for stones,namely that I tried to write a piece of code.
Third,I draw a draft like this to check the correctness of my code.

Code:

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        self.p=p
        self.q=q
        
        self.res=None
        self.recursion(root)
        return self.res
        
    def recursion(self,root):
        
        if root==None:
            return False
        
            
        left=self.recursion(root.left)
        right=self.recursion(root.right)
        if ((left==True)+(right==True)+((root==self.p) or (root==self.q)))==2:
            self.res=root
            return True
        if root==self.p or root==self.q:
            return True
        if left==True or right==True:
            return True
        return False