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9. Binary Tree Inorder Traversal

easyAsked at Intuit

Return the inorder traversal of a binary tree's nodes' values. Intuit asks this in form-rule contexts — TurboTax stores conditional rules as trees and often needs deterministic left-to-right visitation order.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Intuit loops.

  • Glassdoor (2025-11)— TurboTax engine team screen mention.
  • LeetCode Discuss (2026-Q1)— Intuit phone-screen problem.

Problem

Given the root of a binary tree, return the inorder traversal of its nodes' values. Inorder = left subtree, current, right subtree. The follow-up is to do it iteratively.

Constraints

  • The number of nodes in the tree is in the range [0, 100].
  • -100 <= Node.val <= 100

Examples

Example 1

Input
root = [1,null,2,3]
Output
[1,3,2]

Example 2

Input
root = []
Output
[]

Example 3

Input
root = [1]
Output
[1]

Approaches

1. Recursive (warm-up)

Recurse left, push current, recurse right.

Time
O(n)
Space
O(h) recursion stack
function inorderTraversal(root) {
  const out = [];
  function dfs(node) {
    if (!node) return;
    dfs(node.left);
    out.push(node.val);
    dfs(node.right);
  }
  dfs(root);
  return out;
}

Tradeoff: Clean and obvious. The follow-up is to remove the implicit recursion stack.

2. Iterative with explicit stack (optimal)

Use a stack. Push all left children. Pop, emit, then move to the right subtree and repeat.

Time
O(n)
Space
O(h)
function inorderTraversal(root) {
  const out = [];
  const stack = [];
  let cur = root;
  while (cur || stack.length) {
    while (cur) { stack.push(cur); cur = cur.left; }
    cur = stack.pop();
    out.push(cur.val);
    cur = cur.right;
  }
  return out;
}

Tradeoff: Same time/space asymptotically but avoids stack-overflow on deep trees. Useful when h is large.

Intuit-specific tips

Intuit interviewers may ask you to do BOTH versions and articulate when each is appropriate. Recursive is more readable; iterative is safer for very deep trees (e.g., a degenerate left-skewed tree built from a long sorted insertion sequence). Bonus signal: mention Morris traversal (O(1) extra space) as a deeper-level optimization without coding it.

Common mistakes

  • Confusing inorder with preorder (root, left, right) — produces wrong output.
  • Forgetting the inner while loop in iterative — pushes the same node twice.
  • Not handling null root — recursive base case must check.

Follow-up questions

An interviewer at Intuit may pivot to one of these next:

  • Preorder + postorder iterative.
  • Morris traversal (O(1) extra space).
  • Inorder successor of a node — used in BST iterators.

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Output

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FAQ

Why inorder for a BST?

Inorder traversal of a BST visits nodes in sorted order — a property TurboTax exploits when ordering form questions by their dependency tree.

Recursive vs iterative — which does Intuit prefer?

Iterative gets bonus points because it's safer at scale. But have both ready; some interviewers grade on recursive first then ask 'now without recursion'.

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