[LeetCode] 428. Serialize and Deserialize N-ary Tr

iamyoung001 发布于2019-08-16 13:30 / 1895人阅读

Problem

Serialization is the process of converting a data structure or object into a sequence of bits so that it can be stored in a file or memory buffer, or transmitted across a network connection link to be reconstructed later in the same or another computer environment.

Design an algorithm to serialize and deserialize an N-ary tree. An N-ary tree is a rooted tree in which each node has no more than N children. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that an N-ary tree can be serialized to a string and this string can be deserialized to the original tree structure.

For example, you may serialize the following 3-ary tree

as [1 [3[5 6] 2 4]]. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.

Note:

N is in the range of [1, 1000]
Do not use class member/global/static variables to store states. Your serialize and deserialize algorithms should be stateless.

Solution

class Codec {

    // Encodes a tree to a single string.
    public String serialize(Node root) {
        if (root == null) return null;
        List list = new LinkedList<>();
        serializeHelper(root, list);
        return String.join(",", list);
    }
    
    private void serializeHelper(Node root, List list) {
        if (root == null) return;
        list.add(String.valueOf(root.val));
        list.add(String.valueOf(root.children.size()));
        for (Node child: root.children) {
            serializeHelper(child, list);
        }
    }

    // Decodes your encoded data to tree.
    public Node deserialize(String data) {
        if (data == null || data.length() == 0) return null;
        String[] values = data.split(",");
        Queue queue = new LinkedList<>(Arrays.asList(values));
        return deserializeHelper(queue);
    }
    
    private Node deserializeHelper(Queue queue) {
        Node root = new Node();
        root.val = Integer.valueOf(queue.poll());
        int size = Integer.valueOf(queue.poll());
        root.children = new ArrayList<>();
        for (int i = 0; i < size; i++) {
            root.children.add(deserializeHelper(queue));
        }
        return root;
    }
}

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