34. 二叉树中和为某一值的路径

参考了csdn中的代码https://blog.csdn.net/u010005281/article/details/79623396
注意把一个列表加到另一个列表中作为另一个列表的元素，一定要这样写list2.append(list1[:])，不然加的是空的，这涉及到Python的可变对象、不可变对象等机制

class Solution:
    def __init__(self):
        self.onePath = []
        self.PathArray = []
    def FindPath(self, root, expectNumber):
        if root is None:
            return self.PathArray
        self.onePath.append(root.val)
        expectNumber -= root.val
        if expectNumber==0 and not root.left and not root.right:
            self.PathArray.append(self.onePath[:])
        elif expectNumber>0:
            self.FindPath(root.left,expectNumber)
            self.FindPath(root.right,expectNumber)
        self.onePath.pop()
        return self.PathArray

# -*- coding:utf-8 -*-
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
# -*- coding:utf-8 -*-
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
        if not root:
            return []
        ret = []
        path = []
        self.Find(root, expectNumber, ret, path)
        # print ret
        # a = [];self.f(a);print a
        return ret

    def Find(self, root, target, ret, path):
        if not root:
            return 
        path.append(root.val)
        isleaf = (root.left is None and root.right is None)
        if isleaf and target == root.val:
            ret.append(path[:])
        if root.left:
            self.Find(root.left, target-root.val, ret, path)
        if root.right:
            self.Find(root.right, target-root.val, ret, path)
        path.pop()