LeetCode 589. N-ary Tree Preorder Traversal

原题链接在这里：https://leetcode.com/problems/n-ary-tree-preorder-traversal/

题目：

Given the root of an n-ary tree, return the preorder traversal of its nodes' values.

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

Example 1:

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

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: [1,2,3,6,7,11,14,4,8,12,5,9,13,10]

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• 0 <= Node.val <= 104
• The height of the n-ary tree is less than or equal to 1000.

Follow up: Recursive solution is trivial, could you do it iteratively?

题解：

Could use top down recursion.

Time Complexity: O(n). n is the number of nodes in the tree.

Space: O(logn). regardless res. stack space.

AC Java:

 1 /*
 2 // Definition for a Node.
 3 class Node {
 4     public int val;
 5     public List<Node> children;
 6 
 7     public Node() {}
 8 
 9     public Node(int _val) {
10         val = _val;
11     }
12 
13     public Node(int _val, List<Node> _children) {
14         val = _val;
15         children = _children;
16     }
17 };
18 */
19 
20 class Solution {
21     public List<Integer> preorder(Node root) {
22         List<Integer> res = new ArrayList<>();
23         if(root == null){
24             return res;
25         }
26         
27         dfs(root, res);
28         return res;
29     }
30     
31     private void dfs(Node root, List<Integer> res){
32         if(root == null){
33             return;
34         }
35         
36         res.add(root.val);
37         for(Node next : root.children){
38             dfs(next, res);
39         }
40     }
41 }

Could also do it iteratively. It is the method 2 in Binary Tree Preorder Traversal.

Time Complexity: O(n).

Space: O(logn). stack space.

AC Java:

 1 /*
 2 // Definition for a Node.
 3 class Node {
 4     public int val;
 5     public List<Node> children;
 6 
 7     public Node() {}
 8 
 9     public Node(int _val) {
10         val = _val;
11     }
12 
13     public Node(int _val, List<Node> _children) {
14         val = _val;
15         children = _children;
16     }
17 };
18 */
19 
20 class Solution {
21     public List<Integer> preorder(Node root) {
22         List<Integer> res = new ArrayList<>();
23         if(root == null){
24             return res;
25         }
26         
27         Stack<Node> stk = new Stack<>();
28         stk.push(root);
29         while(!stk.isEmpty()){
30             Node cur = stk.pop();
31             res.add(cur.val);
32             for(int i = cur.children.size() - 1; i >= 0; i--){
33                 if(cur.children.get(i) != null){
34                     stk.push(cur.children.get(i));
35                 }    
36             }
37         }
38         
39         return res;
40     }
41 }

标签：Node,Preorder,val,ary,Tree,res,null,root,children
来源： https://www.cnblogs.com/Dylan-Java-NYC/p/16414933.html