# 94. Binary Tree Inorder Traversal

Leetcode #: 94Name: Binary Tree Inorder TraversalDifficulty: EasyCompanies asked: Amazon, Apple, Google, LinkedIn, Meta, Microsoft, UberUrl:https://leetcode.com/problems/binary-tree-inorder-traversal/

# Prompt

You’re given the *root* of a binary tree and it’s your task to return all the nodes within the binary tree *inorder*.

# Approach

This problem is testing your knowledge of whether if you know how to code out an *inorder* traversal algorithm for a Binary Tree. The *inorder* traversal algorithm for a binary tree is quite straightforward once you’ve seen it. I’ll be writing the *inorder* traversal algorithm recursively since we’re dealing with a tree and a tree might have more than 1 level in which a recursive algorithm really comes into handy.

For this problem, as we traverse the Binary Tree *inorder*, we also want to append all the node values into an array so we’ll go ahead and instantiate an empty array.

## Base Case

Our base case is going to be checking if the *root* is null which basically means that during our *inorder* traversal algorithm, we ended up at a null node or an empty tree. In that scenario, we just want to simply return.

`if not root:`

return

## Recursive Case

An *inorder* traversal is different from its counterparts (*preorder* & *postorder*) by having our action (in this case, appending to an array) in between the recursive calls of our left & right subtrees respectively. This is because just like how the name of the traversal algorithm implies (*inorder*), we want to append to our array as if it’s literally in order or in sorted order.

`inorder(root.left, arr)`

arr.append(root.val)

inorder(root.right, arr)

# Code

Thank you for reading!

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