请帮编写代码，输出二叉树中从每个叶子结点到根结点的路径。

以下是 Python 代码实现，其中 `TreeNode` 表示二叉树节点，`findPaths` 函数接受一个二叉树根节点参数，返回一个包含各路径的列表：

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

def findPaths(root):
    res = []
    path = []
    dfs(root, res, path)
    return res

def dfs(node, res, path):
    if not node:
        return
    path.append(node.val)
    if not node.left and not node.right:
        res.append(path[::-1])
    dfs(node.left, res, path)
    dfs(node.right, res, path)
    path.pop()

使用示例：

root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
root.right.left = TreeNode(6)
root.right.right = TreeNode(7)

paths = findPaths(root)
for path in paths:
    print(path)

输出结果：

[4, 2, 1]
[5, 2, 1]
[6, 3, 1]
[7, 3, 1]