What is the optimal time complexity of solving Count Good Nodes In Binary Tree?

The optimized solution achieves a time complexity of O(n) by using sorting, mathematical shortcuts, or hash table lookups.

What is the auxiliary space complexity for Count Good Nodes In Binary Tree?

The space complexity is O(1) since we only use a minimal/constant amount of extra memory for tracking variables (or auxiliary space if dynamic structures are allocated).

What are the typical edge cases we must handle in Count Good Nodes In Binary Tree?

Key edge cases include empty collections, negative or large bounds causing integer overflow, arrays with only one element, or inputs where target criteria are not met.

How can we optimize the brute force approach for Count Good Nodes In Binary Tree?

Brute force methods often inspect all possible combinations. We optimize this by sorting the input list to enable binary search, or tracking processed items in a set to avoid redundant nested loop lookups.

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Count Good Nodes In Binary Tree

Learn how to solve the 'Count Good Nodes In Binary Tree' problem. This detailed resource details brute force and optimized approaches.

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Problem Statement

Medium

Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X.

Return the number of good nodes in the binary tree.

The tree is represented as a level-order list. Implement a function goodNodes(root: list) -> int.

Constraints

•The number of nodes in the binary tree is in the range [1, 100000]
•-10000 <= Node.val <= 10000

Examples

Example 1

Input

[3,1,4,3,None,1,5]

Output

Explanation

Root 3 is always good. Node 4 (3<=4, good). Node 3 under node 1 (3<=3, good). Node 5 (3<=4<=5, good). Node 1 is not good (3>1). Node 1 under 4 is not good (4>1). Total: 4 good nodes.

Example 2

Input

[3,3,None,4,2]

Output

Explanation

Root 3 is good. Node 3 (left child, 3<=3, good). Node 4 (3<=3<=4, good). Node 2 is not good (3>2). Total: 3.

Example 3

Input

[1]

Output

Explanation

The root is always a good node.

Need a Hint?

Perform a recursive tree traversal (DFS) or level-order traversal (BFS) using a queue/stack.

Edge Cases to Watch

Empty list or null input variables
Single item lists/arrays
Extremely large input bounds causing integer or stack overflow

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