Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,
              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1
return
[
   [5,4,11,2],
   [5,8,4,5]
]

# 1.
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        if not root:
            return []
        if not root.left and not root.right and sum == root.val:
            return [[root.val]]
        tmp = self.pathSum(root.left, sum-root.val) + self.pathSum(root.right, sum-root.val)
        return [[root.val]+i for i in tmp]
#2.
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        if not root:
            return []
        pl = []
        self.path(root, sum, [], pl)
        return pl

    def path(self, root, sum, onePath, pl):
        if not root.left and not root.right and root.val == sum:
            onePath.append(root.val)
            pl.append(onePath)
        if root.left:
            self.path(root.left, sum-root.val, onePath+[root.val],pl)
        if root.right:
            self.path(root.right, sum-root.val, onePath+[root.val],pl)