2. Inorder Traversal (Left → Root → Right) ⭐
Traverses the left subtree, processes the current node, then traverses the right subtree (produces sorted order in BST).
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Asked in interviews at
Reference implementation (Python)
def inorder(root):
res = []
def dfs(node):
if not node:
return
dfs(node.left)
res.append(node.val)
dfs(node.right)
dfs(root)
return resEmbed this visualization
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<iframe src="/embed/tree-inorder" width="100%" height="380" style="border:1px solid #e6e8ec;border-radius:14px" loading="lazy" title="3D algorithm visualization by CodeWithMunnaX"></iframe>Frequently asked
What is the time and space complexity of 2. Inorder Traversal (Left → Root → Right) ⭐?
2. Inorder Traversal (Left → Root → Right) ⭐ runs in O(n) time and O(h) auxiliary space.
Is the 2. Inorder Traversal (Left → Root → Right) ⭐ visualizer free?
The core 3D visualizer, Web Audio chimes and step intuition are free. A one-time ₹49 Pro Pass unlocks advanced algorithms, large custom arrays, Battle Arena and reel export.
Which companies ask 2. Inorder Traversal (Left → Root → Right) ⭐ in interviews?
2. Inorder Traversal (Left → Root → Right) ⭐ shows up in interviews at Microsoft, Amazon.
How does the 3D visualization of 2. Inorder Traversal (Left → Root → Right) ⭐ work?
Each array value becomes a 3D pillar whose height maps to its value. Comparisons glow amber, swaps flash red, and settled elements lock green while a floating pointer badge tracks the active index — synchronized line-by-line with the Python code.