Is Median of a Binary Search Tree Level easy or hard?

The problem is typically rated Medium. Traversing the tree is straightforward, but correctly organizing values by level and computing medians efficiently requires understanding of DFS, level grouping, and median calculation.

Median of a Binary Search Tree Level Python/Java solution

Most implementations use DFS where the function receives the current node and depth. Values are appended to a list for that depth, then medians are computed from those lists. The same logic works across Python, Java, C++, Go, and TypeScript.

How to solve Median of a Binary Search Tree Level in O(n)?

Pure O(n) time is difficult if you sort values at each level. A practical optimization is maintaining two heaps per level while traversing with DFS. Each insertion costs O(log w), giving O(n log w) overall while avoiding repeated sorting.

What is the best approach for Median of a Binary Search Tree Level?

A depth-first search that groups node values by level is the most common approach. During traversal, store values in arrays keyed by depth and compute the median for each level after traversal. This solution runs in O(n log w) time where w is the maximum width of the tree and uses O(n) space.

Is Median of a Binary Search Tree Level asked at Google/Amazon/Meta?

Tree traversal and level aggregation problems appear frequently in interviews at companies like Amazon, Google, and Meta. Variants involving medians, averages, or statistics per level test knowledge of DFS, BFS, and data structure design.

What data structure is used in Median of a Binary Search Tree Level?

The core data structures are recursion or a stack for DFS traversal, a hash map or array keyed by depth to store level data, and optionally two heaps (max-heap and min-heap) to maintain medians efficiently.

What is the time complexity of Median of a Binary Search Tree Level?

The typical solution runs in O(n log w) time. DFS visits each of the n nodes once, and computing medians requires sorting or heap operations based on the number of nodes at each level, where w represents the maximum level width.

Description Hints 2 Solution Video 11:27

3831. Median of a Binary Search Tree Level

Medium86.4% acceptancePremium elsewhere, free here

You are given the root of a Binary Search Tree (BST) and an integer level.

The root node is at level 0. Each level represents the distance from the root.

Return the median value of all node values present at the given level. If the level does not exist or contains no nodes, return -1.

The median is defined as the middle element after sorting the values at that level in non-decreasing order. If the number of values at that level is even, return the upper median (the larger of the two middle elements after sorting).

Example 1:

Input: root = [4,null,5,null,7], level = 2

Output: 7

Explanation:

The nodes at level = 2 are [7]. The median value is 7.

Example 2:

Input: root = [6,3,8], level = 1

Output: 8

Explanation:

The nodes at level = 1 are [3, 8]. There are two possible median values, so the larger one 8 is the answer.

Example 3:

Input: root = [2,1], level = 2

Output: -1

Explanation:

There is no node present at level = 2, so the answer is -1.

Constraints:

The number of nodes in the tree is in the range [1, 2 * 10⁵].
1 <= Node.val <= 10⁶
0 <= level <= 2 * 10⁵

Topics

Hints open one at a time

Hint 1

Use DFS or BFS to gather the node values at level.

Hint 2

Sort the values at that level; if the list is empty, return -1, otherwise return the upper median.

11:27

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Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[4,null,5,null,7]
2

Input

[6,3,8]
1

Input

[2,1]
2

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