91 - 二叉树中和为某一值的路径

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
        
def findPath(root, n):
    if not root:
        return []
    result = []
    def findPath2(root, path, currentNum):
        currentNum += root.val
        path.append(root)
        # 判断root是否为叶子节点
        flag = root.left == None and root.right == None
        if currentNum == n and flag:
            onepath = []
            for node in path:
                onepath.append(node.val)
            result.append(onepath)
        if currentNum < n:
            if root.left:
                findPath2(root.left, path, currentNum)
            if root.right:
                findPath2(root.right, path, currentNum)
        path.pop()
    findPath2(root, [], 0)
    return result
    
root = TreeNode(1)
left = TreeNode(2)
right = TreeNode(3)
root.left = left
root.right = right

left1 = TreeNode(4)
right1 = TreeNode(5)
right.left = left1
right.right1 = right1

left2 = TreeNode(6)
left1.left = left2

left3 = TreeNode(11)
left.left = left3

'''

            1
        2       3
              4   5
            6
'''
print(findPath(root, 14))

[[1, 2, 11], [1, 3, 4, 6]]