Is Count Univalue Subtrees easy or hard?

Count Univalue Subtrees is considered a medium difficulty problem. The challenge comes from recognizing that postorder traversal is required and designing a recursive function that returns subtree validity while counting results efficiently.

Count Univalue Subtrees Python/Java solution

Python and Java solutions typically implement a recursive postorder DFS function that returns whether a subtree is univalue. A shared counter variable tracks how many valid subtrees are found while traversing the tree once in O(n) time.

How to solve Count Univalue Subtrees in O(n)?

Use a postorder DFS traversal. Process the left and right subtrees first, then determine whether the current node forms a univalue subtree by comparing its value with its children. Return a boolean indicating subtree validity and increment a counter whenever a valid univalue subtree is found.

What is the best approach for Count Univalue Subtrees?

The best approach uses postorder depth-first search. Each recursive call determines whether the current subtree is univalue by checking results from the left and right children. If both subtrees are univalue and their values match the current node, increment the counter. This solution runs in O(n) time and O(h) space.

Is Count Univalue Subtrees asked at Google/Amazon/Meta?

Tree DFS problems similar to Count Univalue Subtrees appear in interviews at companies like Amazon, Google, and Meta. Interviewers use these questions to evaluate recursive thinking, tree traversal skills, and the ability to propagate state during DFS.

What data structure is used in Count Univalue Subtrees?

The problem uses a binary tree as the primary data structure. The algorithm relies on depth-first search traversal, typically implemented with recursion and a counter variable to track valid univalue subtrees.

What is the time complexity of Count Univalue Subtrees?

The optimal solution runs in O(n) time because each node in the binary tree is visited exactly once during the DFS traversal. Space complexity is O(h), where h is the height of the tree due to the recursion stack.

Description Solution Video 10:13

250. Count Univalue Subtrees

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Given the root of a binary tree, return the number of uni-value subtrees.

A uni-value subtree means all nodes of the subtree have the same value.

Example 1:

Input: root = [5,1,5,5,5,null,5]
Output: 4

Example 2:

Input: root = []
Output: 0

Example 3:

Input: root = [5,5,5,5,5,null,5]
Output: 6

Constraints:

The number of the node in the tree will be in the range [0, 1000].
-1000 <= Node.val <= 1000

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10:13

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250. Count Univalue Subtrees

Asked at 4 companies

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