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9. Binary Tree Inorder Traversal

easyAsked at Booking

Inorder-walk a binary tree — Booking checks this to confirm you can flatten a hierarchical destination tree into a search-ready ordering.

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

Problem

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

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

Left, root, right via recursion.

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

Push lefts, pop and visit, then jump to right; avoids recursion-depth risk on deep trees.

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:

Booking-specific tips

Booking grades for safe recursion — call out the iterative variant as the production choice when traversing a deep destination/region tree.

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