LeetCode 428 - Serialize and Deserialize N-ary Tree

Trie Serialization
http://shibaili.blogspot.com/2018/11/428-serialize-and-deserialize-n-ary-tree.html

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:

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

X. Level Order traverse
http://shibaili.blogspot.com/2018/11/428-serialize-and-deserialize-n-ary-tree.html

    public String serialize(Node root) {

        if (root == null) return "";

         

        Queue<Node> que = new LinkedList<>();

        StringBuilder sb = new StringBuilder();

        sb.append(Integer.toString(root.val)).append(",#,");

        que.add(root);

         

        while (!que.isEmpty()) {

            Node node = que.poll();

            for (Node n : node.children) {

                sb.append(Integer.toString(n.val)).append(",");

                que.add(n);

            }

            sb.append("#,");

        }

         

        return sb.toString();

    }

 

    // Decodes your encoded data to tree.

    public Node deserialize(String data) {

        if (data.length() == 0) return null;

        String[] s = data.split(",");

 

        Queue<Node> que = new LinkedList<>();

        Node root = new Node(Integer.parseInt(s[0]), new ArrayList<Node>());

        que.add(root);

        int i = 1;

         

        while (!que.isEmpty()) {

            Node node = que.poll();

            i++;

            while (!s[i].equals("#")) {

                Node c = new Node(Integer.parseInt(s[i]), new ArrayList<>());

                node.children.add(c);

                que.add(c);

                i++;

            }

        }

         

        return root;

    }

X. Pre-order
https://github.com/Cee/Leetcode/blob/master/428%20-%20Serialize%20and%20Deserialize%20N-ary%20Tree.java

    public String serialize(Node root) {
        if (root == null) { return ""; }
        StringBuilder sb = new StringBuilder();
        sb.append(root.val);
        sb.append(",");
        sb.append(root.children.size());
        sb.append("#");
        for (Node c: root.children) {
            sb.append(serialize(c));
        }
        return sb.toString();
    }

    class Rep {
        Node n;
        int end;
    }
    // Decodes your encoded data to tree.
    public Node deserialize(String data) {
        if (data.length() == 0) { return null; }
        return parse(data, 0, data.length() - 1).n;
    }
    
    private Rep parse(String data, int start, int end) {
        if (start >= end) { return null; }
        int comma = data.indexOf(",", start);
        int split = data.indexOf("#", start);
        int val = Integer.valueOf(data.substring(start, comma));
        int childNum = Integer.valueOf(data.substring(comma + 1, split));
        Node n = new Node();
        n.val = val;
        List<Node> children = new ArrayList<>();
        int last = split;
        for (int i = 0; i < childNum; i++) {
            Rep rep = parse(data, last + 1, end);
            last = rep.end;
            children.add(rep.n);
        }
        n.children = children;
        Rep r = new Rep();
        r.n = n;
        r.end = last;
        return r;
    }

https://segmentfault.com/a/1190000016803936

https://www.twblogs.net/a/5bd1873d2b71776a052c9db6

Serialize and Deserialize an N-ary Tree - GeeksforGeeks
Given an N-ary tree where every node has at-most N children. How to serialize and deserialze it? Serialization is to store tree in a file so that it can be later restored. The structure of tree must be maintained. Deserialization is reading tree back from file.

In an N-ary tree, there are no designated left and right children. An N-ary tree is represented by storing an array or list of child pointers with every node.
The idea is to store an ‘end of children’ marker with every node. The following diagram shows serialization where ‘)’ is used as end of children marker. The diagram is taken from here.
#define MARKER ')'

void serialize(Node *root, FILE *fp)

{

    // Base case

    if (root == NULL) return;

    // Else, store current node and recur for its children

    fprintf(fp, "%c ", root->key);

    for (int i = 0; i < N && root->child[i]; i++)

         serialize(root->child[i],  fp);

    // Store marker at the end of children

    fprintf(fp, "%c ", MARKER);

}

// This function constructs N-ary tree from a file pointed by 'fp'.

// This functionr returns 0 to indicate that the next item is a valid

// tree key. Else returns 0

int deSerialize(Node *&root, FILE *fp)

{

    // Read next item from file. If theere are no more items or next

    // item is marker, then return 1 to indicate same

    char val;

    if ( !fscanf(fp, "%c ", &val) || val == MARKER )

       return 1;

    // Else create node with this item and recur for children

    root = newNode(val);

    for (int i = 0; i < N; i++)

      if (deSerialize(root->child[i], fp))

         break;

    // Finally return 0 for successful finish

    return 0;

}




I guess you don't care about "which" child it is...You have an arbitrary number of children and that's it. If you care about index of the child then yes you cannot know.
https://gist.github.com/ajaynitt/ce01da0d3cfdd5c8db1c
https://mameko.gitbooks.io/pan-s-algorithm-notes/content/26-tree/serilization-and-deserialize-n-ary-tree.html

    public String serialize(NAryNode root) {
        if (root == null) {
            return "";
        }

        StringBuilder sb = new StringBuilder();
        dfsHelper(sb, root);

        return sb.toString();
    }

    private void dfsHelper(StringBuilder sb, NAryNode root) {
        if (root == null) {
            return;
        }

//        sb.append("(");// 可以省一些空间
        sb.append(root.label);
        for (NAryNode child : root.children) {
            dfsHelper(sb, child);
        }
        sb.append(")");
    }

    int index = 0;
    public NAryNode deserialize(String input) {
        if (input == null || input.isEmpty() || index >= input.length()) {
            return null;
        }

//        if (input.charAt(index) == '(') {//省了空间以后就不用跳过'('了
//            index++;// skip '('
            NAryNode node = new NAryNode(input.charAt(index));// create node
            index++;// go to next char
            while (input.charAt(index) != ')') { // if not == ')', has child
                node.children.add(deserialize(input));// so create child node recursively
            }
            index++;// no child / finished creating child go to next char
            return node;
//        }
//        return null;

    }

    public static void main(String[] args) {
        SerializeDeserializeNaryTree sol = new SerializeDeserializeNaryTree();

        NAryNode a = new NAryNode('a');
        NAryNode b = new NAryNode('b');
        NAryNode c = new NAryNode('c');
        NAryNode d = new NAryNode('d');
        NAryNode e = new NAryNode('e');
        NAryNode f = new NAryNode('f');
        NAryNode g = new NAryNode('g');
        NAryNode h = new NAryNode('h');
        NAryNode i = new NAryNode('i');
        NAryNode j = new NAryNode('j');
        NAryNode k = new NAryNode('k');

        a.children.add(b);
        a.children.add(c);
        a.children.add(d);

        b.children.add(e);
        b.children.add(f);

        f.children.add(k);

        d.children.add(g);
        d.children.add(h);
        d.children.add(i);
        d.children.add(j);

        String input = sol.serialize(a);
        System.out.println(input);
        NAryNode res = sol.deserialize(input);
        // 省空间的输出：abe)fk)))c)dg)h)i)j)))
        // 不省的输出：(a(b(e)(f(k)))(c)(d(g)(h)(i)(j)))
    }
}

class NAryNode {
    char label;
    List<NAryNode> children;}




http://ideone.com/TO4dEs

http://eli.thegreenplace.net/2011/09/29/an-interesting-tree-serialization-algorithm-from-dwarf
http://interviewpractise.blogspot.com/2014/11/serialize-and-deserialize-n-ary-tree.html
Related: http://massivealgorithms.blogspot.com/2014/09/serializationdeserialization-of-binary.html
Read full article from Serialize and Deserialize an N-ary Tree - GeeksforGeeks