Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

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]]

# Definition for a binary tree node.class TreeNode:    def __init__(self, x):        self.val = x        self.left = None        self.right = Noneclass Solution:    def pathSum(self, root, sum):        """        :type root: TreeNode        :type sum: int        :rtype: List[List[int]]        """        res = []        if root is None:            return res        def findpath(a,root,sum):            # a.append(root.val)            a = a + [root.val]            if root.left is None and root.right is None and root.val==sum:                res.append(a)                # print('a:',a)                # print('res:',res)            if root.left:                findpath(a,root.left,sum-root.val)            if root.right:                findpath(a,root.right,sum-root.val)        findpath([],root,sum)        return res

使用append实际是修改一个列表，使用+实际是创建一个新的列表。

参考：