二叉树中和为某一值的路径

# -*- coding:utf-8 -*-

import copy
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # 返回二维列表，内部每个列表表示找到的路径
    def FindPath(self, root, expectNumber):
        # write code here
        result=[]
        if not root:
            return result
        
        def FindPathCore(root, path, curNumber):
            curNumber += root.val
            path.append(root.val)
            if curNumber == expectNumber and not root.left and not root.right:
                # onepath=[]
                # for item in path:
                #     onepath.append(item)
                # result.append(onepath)
                result.append(copy.deepcopy(path))
            if curNumber < expectNumber:
                if root.left:
                    FindPathCore(root.left, path, curNumber)
                if root.right:
                    FindPathCore(root.right, path, curNumber)
            path.pop()
        
        path = []
        curNumber = 0
        FindPathCore(root, path, curNumber)
        return result