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Binary TreesEasyLeetCode #700

10. Search in BST ⭐

Efficient logarithmic search leveraging BST invariant (left < root < right) discarding half the search space at each step.

Time O(log n) Space O(1)
Open in 3D Studio

Live loop — open the studio to scrub, add your own array, or race it in Battle Arena.

Asked in interviews at

GoogleAmazon

Reference implementation (Python)

def search_bst(root, val):
    curr = root
    while curr:
        if curr.val == val:
            return curr
        elif val < curr.val:
            curr = curr.left
        else:
            curr = curr.right
    return None

Embed this visualization

Drop a live, looping 10. Search in BST ⭐ canvas into a blog post or course page.

<iframe src="/embed/bst-search" 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 10. Search in BST ⭐?

10. Search in BST ⭐ runs in O(log n) time and O(1) auxiliary space.

Is the 10. Search in BST ⭐ 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 10. Search in BST ⭐ in interviews?

10. Search in BST ⭐ shows up in interviews at Google, Amazon.

How does the 3D visualization of 10. Search in BST ⭐ 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.

More Binary Trees

1. Preorder Traversal (Root → Left → Right) ⭐
O(n)
2. Inorder Traversal (Left → Root → Right) ⭐
O(n)
3. Postorder Traversal (Left → Right → Root) ⭐
O(n)
4. Level Order Traversal (BFS) ⭐
O(n)
5. Height / Depth of Binary Tree ⭐
O(n)
6. Count Total Nodes
O(n)