递归与二叉树_leetcode113

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


#coding=utf-8
# 解题思路： 递归返回值 20190305 找工作期间
# 递归返回值的处理、要定义好函数的返回逻辑或状态或者值，并从底向上构建
class Solution(object):
 def pathSum(self, root, sum):
 """
 :type root: TreeNode
 :type sum: int
 :rtype: List[List[int]]
 """

 res = []
 if not root:
 return res

 if not root.left and not root.right:
 if root.val == sum:
 res.append(str(root.val))

 return res

 leftPaths = self.pathSum(root.left,sum - root.val)

 for i in range(len(leftPaths)):

 res.append(str(root.val) + "," + leftPaths[i])

 rightPaths = self.pathSum(root.right,sum - root.val)

 for i in range(len(rightPaths)):

 res.append(str(root.val) + "," + rightPaths[i])

 return res


class Solution2(object):
 def pathSum(self, root, sum):
 """
 :type root: TreeNode
 :type sum: int
 :rtype: List[List[int]]
 """

 res = []
 if not root:
 return res

 if not root.left and not root.right:
 if root.val == sum:
 res.append([root.val])

 return res

 leftPaths = self.pathSum(root.left,sum - root.val)

 for i in range(len(leftPaths)):

 res.append([root.val] + leftPaths[i])

 rightPaths = self.pathSum(root.right,sum - root.val)


 for i in range(len(rightPaths)):

 res.append([root.val] + rightPaths[i])

 return res