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9. Binary Tree Inorder Traversal

easyAsked at Glassdoor

Return the inorder traversal of a binary tree — Glassdoor uses this to gauge tree recursion fluency, then probes the iterative version.

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

Problem

Given the root of a binary tree, return the inorder traversal of its nodes' values. Inorder traversal visits left subtree, then node, then right subtree.

Constraints

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

Examples

Example 1

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

Example 2

Input
root = []
Output
[]

Approaches

1. Recursive DFS

Recurse left, push value, recurse right.

Time
O(n)
Space
O(h)
function inorder(root, out = []) {
  if (!root) return out;
  inorder(root.left, out);
  out.push(root.val);
  inorder(root.right, out);
  return out;
}

Tradeoff:

2. Iterative with stack

Walk left as far as possible, pop, then move right; explicit stack avoids recursion limits.

Time
O(n)
Space
O(h)
function inorderIter(root) {
  const out = [], 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:

Glassdoor-specific tips

Glassdoor often follows the recursive answer with 'now show me iterative' — bonus signal is mentioning Morris traversal for O(1) space, which echoes their tight-memory aggregation jobs.

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