代码一

用list或者字符串方式返回二叉树的所有路径：https://www.cnblogs.com/panweiwei/p/13752895.html

class Solution(object):
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        if not root:
            return []
        res = []
        road = []
        self.pre_order(root, sum, res, road)
        return res

    def pre_order(self, node, total, res, road):
        # 当前节点为空，直接return
        if not node:
            return
        # 否则将当前节点加入路径中
        road.append(node.val)
        # 目标值减掉当前节点的值
        total -= node.val
        # 当前节点是叶子、且和为目标值，则将路径加入外层list中
        if total == 0 and not node.left and not node.right:
            res.append(list(road))
        # 前序式递归当前节点的左右子树
        self.pre_order(node.left, total, res, road)
        self.pre_order(node.right, total, res, road)
        # 每趟递归前将当前节点pop()
        road.pop()
        # 返回外层list
        return res

代码二

class Solution(object):
    def pathSum(self, root, sumt):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        ans = []
        if not root:
            return ans
        stack = [(root, [root.val])]
        while stack:
            node, temp = stack.pop()
            if not node.left and not node.right and sum(temp) == sumt:
                ans.append(temp)
            if node.left:
                stack.append((node.left, temp + [node.left.val]))
            if node.right:
                stack.append((node.right, temp + [node.right.val]))
        return ans[::-1]