Leetcode #2792: Count Nodes That Are Great Enough
In this guide, we solve Leetcode #2792 Count Nodes That Are Great Enough in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Tree, Depth-First Search, Divide and Conquer, Binary Tree
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: root = [7,6,5,4,3,2,1], k = 2
Output: 3
Explanation: Number the nodes from 1 to 7.
The values in the subtree of node 1: {1,2,3,4,5,6,7}. Since node.val == 7, there are 6 nodes having a smaller value than its value. So it's great enough.
The values in the subtree of node 2: {3,4,6}. Since node.val == 6, there are 2 nodes having a smaller value than its value. So it's great enough.
The values in the subtree of node 3: {1,2,5}. Since node.val == 5, there are 2 nodes having a smaller value than its value. So it's great enough.
It can be shown that other nodes are not great enough.
See the picture below for a better understanding.
Python Solution
# 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
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.