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