binary-tree-inorder-traversal

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Algorithms

Given the <code>root</code> of a binary tree, return the inorder traversal of its nodes&#39; values.

&nbsp;
Example 1:

<div class="example-block">
Input: root = [1,null,2,3]

Output: [1,3,2]

Explanation:

<img alt="" src="https://assets.leetcode.com/uploads/2024/08/29/screenshot-2024-08-29-202743.png" style="width: 200px; height: 264px;" />
</div>

Example 2:

<div class="example-block">
Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [4,2,6,5,7,1,3,9,8]

Explanation:

<img alt="" src="https://assets.leetcode.com/uploads/2024/08/29/tree_2.png" style="width: 350px; height: 286px;" />
</div>

Example 3:

<div class="example-block">
Input: root = []

Output: []
</div>

Example 4:

<div class="example-block">
Input: root = [1]

Output: [1]
</div>

&nbsp;
Constraints:

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 100]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

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

inorderTraversal

Binary Tree Inorder Traversal

Validate Binary Search Tree

validate-binary-search-tree

验证二叉搜索树

Binary Tree Preorder Traversal

binary-tree-preorder-traversal

二叉树的前序遍历

Binary Tree Postorder Traversal

binary-tree-postorder-traversal

二叉树的后序遍历

Binary Search Tree Iterator

binary-search-tree-iterator

二叉搜索树迭代器

Kth Smallest Element in a BST

kth-smallest-element-in-a-bst

二叉搜索树中第 K 小的元素

Closest Binary Search Tree Value II

closest-binary-search-tree-value-ii

最接近的二叉搜索树值 II

Inorder Successor in BST

inorder-successor-in-bst

二叉搜索树中的中序后继

Convert Binary Search Tree to Sorted Doubly Linked List

convert-binary-search-tree-to-sorted-doubly-linked-list

将二叉搜索树转化为排序的双向链表

Minimum Distance Between BST Nodes

minimum-distance-between-bst-nodes

二叉搜索树节点最小距离

Stack

Tree

Depth-First Search

Binary Tree

给定一个二叉树的根节点 <code>root</code> ，返回 它的 中序&nbsp;遍历 。

&nbsp;

示例 1：
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/15/inorder_1.jpg" style="height: 200px; width: 125px;" />
<pre>
输入：root = [1,null,2,3]
输出：[1,3,2]
</pre>

示例 2：

<pre>
输入：root = []
输出：[]
</pre>

示例 3：

<pre>
输入：root = [1]
输出：[1]
</pre>

&nbsp;

提示：

<ul>
	<li>树中节点数目在范围 <code>[0, 100]</code> 内</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
</ul>

&nbsp;

进阶:&nbsp;递归算法很简单，你可以通过迭代算法完成吗？