题解 | #二叉树中和为某一值的路径(一)#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param root TreeNode类 
# @param sum int整型 
# @return bool布尔型
#
class Solution:
    def hasPathSum(self , root: TreeNode, sum: int) -> bool:
        # write code here
        res, path = [], []
        def recur(root, tar):
            if not root: return
            path.append(root.val)
            tar -= root.val
            if tar == 0 and not root.left and not root.right:
                res.append(list(path))
            recur(root.left, tar)
            recur(root.right, tar)
            path.pop()

        recur(root, sum)
        return res