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9. Binary Tree Inorder Traversal

easyAsked at DigitalOcean

Return the inorder traversal of a binary tree — DigitalOcean uses this to test tree-walk fluency that mirrors traversing a region/zone/rack hierarchy for billing rollups.

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 means left subtree, node, right subtree.

Constraints

  • 0 <= number of nodes <= 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 left, push node, DFS 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 stack

Use a stack: push lefts, pop and emit, then move to right child.

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

Tradeoff:

DigitalOcean-specific tips

DigitalOcean prefers the iterative version because their infra topology graphs can be deep enough to blow recursion stacks during multi-region traversal.

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Output

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