递归遍历二叉树--python

#输入两颗二叉树A、B，判断B是不是A的子结构
#思路：递归遍历二叉数A，找到A中和B根节点相同的节点，接着顺着这个相等的节点判断子结构是不是和B树的子结构相同
def subTree(pRoot1,pRoot2):
    result = False
    if pRoot1 is not None and pRoot2 is not None:
    	if pRoot1.val == pRoot2.val:
    		result = IsTree1ContainsTree2(pRoot1,pRoot2)
    	if not result:
    		result = subTree(pRoot1.left,pRoot2)
    	if not result:
    		result = subTree(pRoot1.right,pRoot2)
    return result

def IsTree1ContainsTree2(pRoot1, pRoot2):
    if pRoot2 is None:
        return True
    if pRoot1 is None:  # 1已经到了叶子结点，为空了，肯定不会包含2了
        return False
    if pRoot1.val != pRoot2.val:  # 两个树对应点的值不相等，不用继续了，不包含
        return False
        # 经历过上面的筛选后，这个节点通过了考验，
        # 继续向下，判断各自的左右子节点是否一样
        # 用递归方法可以直接写出
    return IsTree1ContainsTree2(pRoot1.left, pRoot2.left) and IsTree1ContainsTree2(pRoot1.right, pRoot2.right)

root1 = TreeNode(8)
root1.left = TreeNode(8)
root1.right = TreeNode(7)
root1.left.left = TreeNode(9)
root1.left.right = TreeNode(2)
root1.left.right.left = TreeNode(4)
root1.left.right.right = TreeNode(7)

root2 = TreeNode(8)
root2.left = TreeNode(9)
root2.right = TreeNode(1)

# print(subTree(root1,root2))