leetcode讲解--589. N-ary Tree Preorder Traversal

题目Given an n-ary tree, return the preorder traversal of its nodes’ values.For example, given a 3-ary tree:Return its preorder traversal as: [1,3,5,6,2,4].Note:Recursive solution is trivial, could you do it iteratively?题目地址讲解如果用递归来解题的话，还是非常简单的。如果要用迭代来解题，无非就是用栈来实现。要注意的一点是，需要一个额外的栈来把压栈的顺序倒一下序。Java代码递归代码：/// Definition for a Node.class Node { public int val; public List<Node> children; public Node() {} public Node(int _val,List<Node> _children) { val = _val; children = _children; }};/class Solution { private List<Integer> result = new ArrayList<>(); public List<Integer> preorder(Node root) { if(root==null){ return result; } result.add(root.val); for(Node node:root.children){ preorder(node); } return result; }}迭代代码：/// Definition for a Node.class Node { public int val; public List<Node> children; public Node() {} public Node(int _val,List<Node> _children) { val = _val; children = _children; }};/class Solution { private List<Integer> result = new ArrayList<>(); private Stack<Node> stack = new Stack<>(); private Stack<Node> stack_temp = new Stack(); public List<Integer> preorder(Node root) { if(root==null){ return result; } stack.push(root); while(!stack.empty()){ Node theNode = stack.pop(); result.add(theNode.val); for(Node node:theNode.children){ stack_temp.push(node); } while(!stack_temp.empty()){ stack.push(stack_temp.pop()); } } return result; }}