Count Good Nodes In Binary Tree
Detailed guide and Python implementation for the 'Count Good Nodes In Binary Tree' problem.
1. Concept Overview
The 'Count Good Nodes In Binary Tree' problem is a key challenge in the Trees section.
This implementation focuses on medium-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Count Good Nodes In Binary Tree.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Count Good Nodes In Binary Tree carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
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.
- •The number of nodes in the binary tree is in the range [1, 100000]
- •-10000 <= Node.val <= 10000
Examples
[3,1,4,3,None,1,5]
4
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.
[3,3,None,4,2]
3
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.
[1]
1
The root is always a good node.
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def good_nodes_opt(root):
if isinstance(root, list):
r = build_tree(root)
return good_nodes_opt_helper(r)
return good_nodes_opt_helper(root)
def good_nodes_opt_helper(root: TreeNode) -> int:
def dfs(node, max_val):
if not node:
return 0
res = 1 if node.val >= max_val else 0
max_val = max(max_val, node.val)
res += dfs(node.left, max_val)
res += dfs(node.right, max_val)
return res
return dfs(root, float('-inf'))Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def good_nodes_brute(root):
if isinstance(root, list):
r = build_tree(root)
return good_nodes_brute_helper(r)
return good_nodes_brute_helper(root)
def good_nodes_brute_helper(root: TreeNode) -> int:
def dfs(node, path):
if not node: return 0
is_good = 1 if all(x <= node.val for x in path) else 0
path.append(node.val)
left = dfs(node.left, path)
right = dfs(node.right, path)
path.pop()
return is_good + left + right
return dfs(root, [])Algorithm Pattern Checklist
When dealing with Trees data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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