Substructure tree python offer to prove safety

Title Description

Two binary inputs A, B, B is judged not substructure A's. (Ps: we agreed empty tree is not a tree any substructure)

Thinking

Traversing a first tree, to find the root of the tree, and b value equal to the node, the determination is not to find the subtree.

Judgment is not subtree when the recursive method, the root node is determined, and then determines the left child and right child.

Code

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    def HasSubtree(self, pRoot1, pRoot2):
        # write code here
        result = False
        if pRoot1 and pRoot2:
            if pRoot1.val == pRoot2.val:
                result = self.isSubtree(pRoot1,pRoot2)
            if not result and pRoot1.left:
                result = self.isSubtree(pRoot1.left,pRoot2)
            if not result and pRoot2.right:
                result = self.isSubtree(pRoot1.right,pRoot2)
        return result
    
    def isSubtree(self,p1,p2):
        if not p2:
            return True
        if not p1:
            return False
        if p1.val != p2.val:
            return False
        check_left = self.isSubtree(p1.left, p2.left)
        check_right = self.isSubtree(p1.right,p2.right)
        return check_left and check_right