[LeetCode] 1490. Clone N-ary Tree

Given a root of an N-ary tree, return a deep copy (clone) of the tree.

Each node in the n-ary tree contains a val (int) and a list (List[Node]) of its children.

class Node {
    public int val;
    public List<Node> children;
}

Nary-Tree input serialization is represented in their level order traversal, each group of children is separated by the null value (See examples).

Follow up: Can your solution work for the graph problem?

Example 1:

Input: root = [1,null,3,2,4,null,5,6]
Output: [1,null,3,2,4,null,5,6]

Example 2:

Input: root = [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]
Output: [1,null,2,3,4,5,null,null,6,7,null,8,null,9,10,null,null,11,null,12,null,13,null,null,14]

Constraints:

The depth of the n-ary tree is less than or equal to 1000.
The total number of nodes is between [0, 10^4].

克隆N叉树。题目就是题意。这道题有两种做法，BFS和DFS，两种做法的时间和空间复杂度都是O(n)。建议可以先做一下133题，熟悉一下克隆是怎么一回事。

DFS

对递归有感觉的同学会发现DFS的思路很容易做。首先自然是clone一下当前遇到的节点，得到copy节点；对于当前节点的所有孩子，不光需要为这每一个孩子clone一份，同时也要把这些clone的孩子放到copy节点的children指针上。

 1 class Solution {
 2     public Node cloneTree(Node root) {
 3         // corner case
 4         if (root == null) {
 5             return null;
 6         }
 7         // normal case
 8         Node copy = new Node(root.val);
 9         for (Node child : root.children) {
10             copy.children.add(cloneTree(child));
11         }
12         return copy;
13     }
14 }

BFS

既然是BFS，肯定是需要用到一个队列queue去遍历所有的节点。遍历的时候，需要为每一个遍历到的节点clone一份，并用hashmap记录原生版本和复制版本的对应关系；同时，在遍历每个节点的children们的时候，需要clone每一个孩子节点，用hashmap记录原生版本和复制版本的对应关系，同时需要把clone的孩子加到当前节点的clone的children指针上。

 1 class Solution {
 2     public Node cloneTree(Node root) {
 3         // corner case
 4         if (root == null) {
 5             return null;
 6         }
 7         // normal case
 8         HashMap<Node, Node> map = new HashMap<>();
 9         Queue<Node> queue = new LinkedList<>();
10         queue.offer(root);
11         while (!queue.isEmpty()) {
12             int size = queue.size();
13             for (int i = 0; i < size; i++) {
14                 // 复制当前节点并记录clone关系
15                 Node cur = queue.poll();
16                 Node curCopy = map.getOrDefault(cur, new Node(cur.val));
17                 map.put(cur, curCopy);
18                 // 遍历孩子节点
19                 for (Node child : cur.children) {
20                     queue.offer(child);
21                     Node childCopy = map.getOrDefault(child, new Node(child.val));
22                     curCopy.children.add(childCopy);
23                     map.put(child, childCopy);
24                 }
25             }
26         }
27         return map.get(root);
28     }
29 }