None in Python not equivalent to null in OJ

  • 0

    I have written a simple code to solve the problem (nothing too fancy) in Python. I have attached my code at the end.
    According to the OJ, for the input [1,null,2], the expected output is [1,null,2] which is correct.

    My output is coming out to be [1,None,2] which apparently doesn't match with the OJ. I thought None in Python and null in OJ were supposed to be equivalent. What am I doing wrong and why isn't the OJ not parsing None as null?

    Any help would be appreciated.

    # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    class Codec:
        def serialize(self, root):
            """Encodes a tree to a single string.
            :type root: TreeNode
            :rtype: str
            li, que = [], []
            if root: que.append(root)
            while len(que) > 0:
                root = que.pop(0)
                if not root:
            return ",".join(map(str, li))
        def deserialize(self, data):
            """Decodes your encoded data to tree.
            :type data: str
            :rtype: TreeNode
            if not data:
                return None
            li = map(lambda v: int(v) if v != "null" else None, data.split(","))
            while li and li[-1] is None: li.pop()
            # print li
            idx = 0
            root = TreeNode(li[idx])
            que = [(root, idx)]
            while len(que) > 0:
                curr, idx = que.pop(0)
                # print "hola", curr, idx,
                if not curr:
                    # print
                # print "val", curr.val,
                if 2*idx+1 < len(li):
                    curr.left = TreeNode(li[2*idx + 1])
                    que.append((curr.left, 2*idx+1))
                if 2*idx+2 < len(li):
                    curr.right = TreeNode(li[2*idx + 2])
                    que.append((curr.right, 2*idx+2))
                # print "que", que
            return root
    # Your Codec object will be instantiated and called as such:
    # codec = Codec()
    # codec.deserialize(codec.serialize(root))

  • 0

    For that example, root.left should be None. Not a node with a val attribute of None, which is what you're producing.

  • 0

    @StefanPochmann Yes, I realised as much. There are a lot of issues here with falsy types alone. Thanks for replying!

Log in to reply

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.