二叉樹中和為某一值的路徑

class Solution:
    def FindPath(self, root, expectNumber):
        if root ==None:
            return []
        elif root.left == None and root.right == None and root.val == expectNumber:
            return [[root.val]]
        res=[]
        left=self.FindPath(root.left, expectNumber-root.val)
        right=self.FindPath(root.right, expectNumber-root.val)
        for item in left+right:
            res.append([root.val]+item)
        return res