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9. Binary Tree Inorder Traversal

easyAsked at Slack

Return the inorder traversal of a binary tree's values.

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

Problem

Given the root of a binary tree, return the inorder traversal of its nodes' values (left, root, right).

Constraints

  • Number of nodes 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

Simple recursion: left, push value, right.

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

Tradeoff:

2. Iterative with stack

Push lefts onto a stack, then pop, record, and move right. Handles deep trees without recursion limit.

Time
O(n)
Space
O(h)
function inorderTraversal(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:

Slack-specific tips

Slack channel-tree (workspaces > channels > threads) is a real recursive structure — they may ask you to convert recursion to iteration to avoid stack overflow on huge workspaces.

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Output

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