题目链接

题目描述

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

题解

前序遍历，把当前节点的子节点依次添加到一个栈中，依次遍历就好

代码

/*// Definition for a Node.class Node {    public int val;    public List
    
      children;    public Node() {}    public Node(int _val,List
     
       _children) {        val = _val;        children = _children;    }};*/class Solution {    public List
      
        preorder(Node root) {        List
       
         list = new ArrayList<>();        if (root == null) { return list; }        Stack
        
          s = new Stack<>();        s.add(root);        while (!s.isEmpty()) {            Node node = s.pop();            list.add(node.val);            for (int i = node.children.size() - 1; i>=0; i--) {                s.add(node.children.get(i));            }        }        return list;    }}