binary-tree-level-order-traversal

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Algorithms

Given the <code>root</code> of a binary tree, return the level order traversal of its nodes&#39; values. (i.e., from left to right, level by level).

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg" style="width: 277px; height: 302px;" />
<pre>
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]
</pre>

Example 2:

<pre>
Input: root = [1]
Output: [[1]]
</pre>

Example 3:

<pre>
Input: root = []
Output: []
</pre>

&nbsp;
Constraints:

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 2000]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
</ul>

levelOrder

Binary Tree Level Order Traversal

Binary Tree Zigzag Level Order Traversal

binary-tree-zigzag-level-order-traversal

二叉树的锯齿形层序遍历

Binary Tree Level Order Traversal II

binary-tree-level-order-traversal-ii

二叉树的层序遍历 II

Minimum Depth of Binary Tree

minimum-depth-of-binary-tree

二叉树的最小深度

Binary Tree Vertical Order Traversal

binary-tree-vertical-order-traversal

二叉树的垂直遍历

Average of Levels in Binary Tree

average-of-levels-in-binary-tree

二叉树的层平均值

N-ary Tree Level Order Traversal

n-ary-tree-level-order-traversal

N 叉树的层序遍历

Cousins in Binary Tree

cousins-in-binary-tree

二叉树的堂兄弟节点

Minimum Number of Operations to Sort a Binary Tree by Level

minimum-number-of-operations-to-sort-a-binary-tree-by-level

逐层排序二叉树所需的最少操作数目

Divide Nodes Into the Maximum Number of Groups

divide-nodes-into-the-maximum-number-of-groups

将节点分成尽可能多的组

Tree

Breadth-First Search

Binary Tree

给你二叉树的根节点 <code>root</code> ，返回其节点值的 层序遍历 。 （即逐层地，从左到右访问所有节点）。

&nbsp;

示例 1：
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/tree1.jpg" style="width: 277px; height: 302px;" />
<pre>
输入：root = [3,9,20,null,null,15,7]
输出：[[3],[9,20],[15,7]]
</pre>

示例 2：

<pre>
输入：root = [1]
输出：[[1]]
</pre>

示例 3：

<pre>
输入：root = []
输出：[]
</pre>

&nbsp;

提示：

<ul>
	<li>树中节点数目在范围 <code>[0, 2000]</code> 内</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
</ul>