n-ary-tree-preorder-traversal

java

python

python3

mysql

mssql

oraclesql

csharp

javascript

typescript

bash

swift

kotlin

dart

golang

ruby

scala

rust

racket

erlang

elixir

pythondata

postgresql

cangjie

Algorithms

Given the <code>root</code> of an n-ary tree, return the preorder traversal of its nodes&#39; values.

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

&nbsp;
Example 1:

<img src="https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png" style="width: 100%; max-width: 300px;" />

<pre>
Input: root = [1,null,3,2,4,null,5,6]
Output: [1,3,5,6,2,4]
</pre>

Example 2:

<img alt="" src="https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png" style="width: 296px; height: 241px;" />

<pre>
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]
</pre>

&nbsp;
Constraints:

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 104]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 104</code></li>
	<li>The height of the n-ary tree is less than or equal to <code>1000</code>.</li>
</ul>

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

preorder

N-ary Tree Preorder Traversal

Binary Tree Preorder Traversal

binary-tree-preorder-traversal

二叉树的前序遍历

N-ary Tree Level Order Traversal

n-ary-tree-level-order-traversal

N 叉树的层序遍历

N-ary Tree Postorder Traversal

n-ary-tree-postorder-traversal

N 叉树的后序遍历

Stack

Tree

Depth-First Search

给定一个 n&nbsp;叉树的根节点 <meta charset="UTF-8" />&nbsp;<code>root</code>&nbsp;，返回 其节点值的 前序遍历 。

n 叉树 在输入中按层序遍历进行序列化表示，每组子节点由空值 <code>null</code> 分隔（请参见示例）。

 
示例 1：

<img src="https://assets.leetcode.com/uploads/2018/10/12/narytreeexample.png" style="height: 193px; width: 300px;" />

<pre>
输入：root = [1,null,3,2,4,null,5,6]
输出：[1,3,5,6,2,4]
</pre>

示例 2：

<img alt="" src="https://assets.leetcode.com/uploads/2019/11/08/sample_4_964.png" style="height: 272px; width: 300px;" />

<pre>
输入：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]
输出：[1,2,3,6,7,11,14,4,8,12,5,9,13,10]
</pre>

&nbsp;

提示：

<ul>
	<li>节点总数在范围<meta charset="UTF-8" />&nbsp;<code>[0, 104]</code>内</li>
	<li><code>0 &lt;= Node.val &lt;= 104</code></li>
	<li>n 叉树的高度小于或等于 <code>1000</code></li>
</ul>

&nbsp;

进阶：递归法很简单，你可以使用迭代法完成此题吗?