Leetcode 113. Path Sum II

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]
]

 class Solution(object):
     def pathSum(self, root, sum):
         """
         :type root: TreeNode
         :type sum: int
         :rtype: List[List[int]]
         """
         res = []
         self.helper(root, sum, res, [])

         return res

     def helper(self, root, sum, res, path):
         if not root:
             return 

         if not root.left and not root.right and root.val == sum:
             path.append(root.val)
             res.append([x for x in path])
             return 

         self.helper(root.left, sum - root.val, res, path + [root.val])
         self.helper(root.right, sum - root.val, res, path + [root.val])
         return 

上面的解法由于在L22和L23中没有重新定义path，所以可以不用pop。如果重新定义path，可用如下解法。这二者的区别类似 Binary tree path 一题中 九章算法和书影的给出的解法的区别。

 class Solution(object):
     def pathSum(self, root, sum):
         """
         :type root: TreeNode
         :type sum: int
         :rtype: List[List[int]]
         """
         res = []
         self.helper(root, sum, res, [])

         return res

     def helper(self, root, sum, res, path):
         if not root:
             return 

         if not root.left and not root.right and root.val == sum:
             path.append(root.val)
             res.append([x for x in path])
             return 

         self.helper(root.left, sum - root.val, res, path + [root.val])
         self.helper(root.right, sum - root.val, res, path + [root.val])
         return