How hard is LeetCode 2792. Count Nodes That Are Great Enough?

LeetCode 2792. Count Nodes That Are Great Enough is rated Hard on LeetCode.

What is the time complexity of LeetCode 2792. Count Nodes That Are Great Enough?

The Python solution on this page runs in O(V+E).

What is the space complexity of LeetCode 2792. Count Nodes That Are Great Enough?

The Python solution on this page uses O(V) auxiliary space.

What topics does LeetCode 2792. Count Nodes That Are Great Enough cover?

LeetCode 2792. Count Nodes That Are Great Enough is tagged Tree, Depth-First Search, Divide and Conquer and Binary Tree on LeetCode.

Is LeetCode 2792. Count Nodes That Are Great Enough a premium problem?

Yes. LeetCode 2792. Count Nodes That Are Great Enough is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

Count Nodes That Are Great Enough — LeetCode 2792 Python Solution

HardLeetCode PremiumTreeDepth-First SearchDivide and ConquerBinary Tree

Problem: #2792
Pattern: Tree Traversal
Reading time: 5 min
Source: leetcode.com

The problem

You are given a root to a binary tree and an integer k. A node of this tree is called great enough if the followings hold: Its subtree has at least k nodes.

Example

Input: root = [7,6,5,4,3,2,1], k = 2
Output: 3
Explanation: Number the nodes from 1 to 7.

Python solution

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def countGreatEnoughNodes(self, root: Optional[TreeNode], k: int) -> int:
        def push(pq, x):
            heappush(pq, x)
            if len(pq) > k:
                heappop(pq)

        def dfs(root):
            if root is None:
                return []
            l, r = dfs(root.left), dfs(root.right)
            for x in r:
                push(l, x)
            if len(l) == k and -l[0] < root.val:
                nonlocal ans
                ans += 1
            push(l, -root.val)
            return l

        ans = 0
        dfs(root)
        return ans

Complexity

Measure	Complexity
Time	O(V+E)
Space	O(V) auxiliary

Pattern: Tree Traversal

Choose the order — preorder, inorder, postorder, level — and the problem solves itself. LeetCode 2792. Count Nodes That Are Great Enough is filed here because LeetCode tags it Tree and Binary Tree, which is the vocabulary this hub collects.

The tree traversal guide has the Python template for the pattern and the 225 LeetCode problems that use it.

Frequently asked questions

How hard is LeetCode 2792. Count Nodes That Are Great Enough?: LeetCode 2792. Count Nodes That Are Great Enough is rated Hard on LeetCode.
What is the time complexity of LeetCode 2792. Count Nodes That Are Great Enough?: The Python solution on this page runs in O(V+E).
What is the space complexity of LeetCode 2792. Count Nodes That Are Great Enough?: The Python solution on this page uses O(V) auxiliary space.
What topics does LeetCode 2792. Count Nodes That Are Great Enough cover?: LeetCode 2792. Count Nodes That Are Great Enough is tagged Tree, Depth-First Search, Divide and Conquer and Binary Tree on LeetCode.
Is LeetCode 2792. Count Nodes That Are Great Enough a premium problem?: Yes. LeetCode 2792. Count Nodes That Are Great Enough is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.