DSA interview question

Binary Tree Inorder Traversal

Easy · Data structures · Topics: tree, dfs, recursion

The problem

A contacts app stores names in a binary search tree, and its "show all, sorted" button walks the tree inorder: left subtree, then the node, then the right subtree. For a BST that visits the keys in sorted order, and for any binary tree it gives the natural "left to right" reading of the nodes. The recursive version mirrors the definition directly.

Example: [1, null, 2, 3] → [1, 3, 2].

Function signature

vector<int> inorderTraversal(TreeNode* root);

Test cases

Input (level order)	Output	Explanation
`[1, null, 2, 3]`	`[1, 3, 2]`
`[1, 2, 3, 4, 5, null, 8, null, null, 6, 7, 9]`	`[4, 2, 6, 5, 7, 1, 3, 9, 8]`
`[2, 1, 3]`	`[1, 2, 3]`	A BST comes out sorted
`[]`	`[]`
`[1]`	`[1]`

(Confirm: LeetCode's level-order input format; the tree may be empty.)

How interviewers phrase it

"Given the root of a binary tree, return the inorder traversal of its nodes' values" (LeetCode 94), often followed by "recursive is trivial, can you do it iteratively?"

Clarifying questions to ask

Recursive or iterative?
Can the tree be empty or very deep?
Is it a BST? (Then inorder means sorted order, which many BST problems rely on.)

Follow-ups to expect

"Why does inorder sort a BST?"
"Iterative version?"
"Morris inorder?"
"Kth smallest in a BST?"
"Complexity?"

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