What is the time complexity of Count Good Nodes in Binary Tree?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count Good Nodes in Binary Tree?

This problem uses Tree, Depth-First Search, Breadth-First Search, Binary Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Good Nodes in Binary Tree?

Insight 1: We need to keep track of the maximum value encountered so far along the path from the root to the current node. Insight 2: Depth-First Search (DFS) is a natural fit for traversing the tree and maintaining the path information.

Which companies ask LeetCode 1448. Count Good Nodes in Binary Tree in interviews?

This problem is frequently asked at Docusign, josh technology.

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Count Good Nodes in Binary Tree - LeetCode 1448 Solution

Count Good Nodes in Binary Tree - Complete Solution Guide

Count Good Nodes in Binary Tree is LeetCode problem 1448, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. Example 1: Input: root = [3,1,4,3,null,1,5] Output: 4 Explanation: Nodes in blue are good . Root Node (3) is always a good node. Node 4 -> (3,4) is the maximum value in the path starting from the root. Node 5 -> (3,4,5) is the maximum value in the path Node 3 -> (3,1,3) is the maximum value in the path. Exa

Detailed Explanation

The problem asks us to traverse a binary tree and count the number of 'good' nodes. A node is considered 'good' if, on the path from the root to that node, there are no nodes with a value greater than the node's value. In simpler terms, the node's value must be the maximum value seen so far along the path from the root. We are given the root node of the binary tree and must return the total number of good nodes.

Solution Approach

The solution uses a Depth-First Search (DFS) traversal of the binary tree. For each node, it checks if the node's value is greater than or equal to the maximum value seen so far on the path from the root. If it is, the node is a 'good' node, and the count is incremented. The maximum value is then updated to be the maximum of the current maximum and the node's value. The DFS continues recursively down the left and right subtrees, passing the updated maximum value to each recursive call.

Step-by-Step Algorithm

Step 1: Initialize a counter to 0 to store the number of good nodes.
Step 2: Define a recursive helper function `dfs(node, max_val)` where `node` is the current node being visited and `max_val` is the maximum value seen from root to `node`.
Step 3: Base case for the recursion: If the current `node` is `None`, return 0 (no good nodes found in an empty subtree).
Step 4: Check if the current `node.val` is greater than or equal to `max_val`. If it is, increment the counter by 1 (it's a good node).
Step 5: Update `max_val` to be the maximum of the current `max_val` and `node.val`.
Step 6: Recursively call `dfs(node.left, max_val)` and add the result to the counter.
Step 7: Recursively call `dfs(node.right, max_val)` and add the result to the counter.
Step 8: Return the final count.
Step 9: Call the `dfs` function starting from the `root` node with an initial `max_val` smaller than any possible node value (e.g., -10001).
Step 10: Return the value returned by the `dfs` function as the result.

Key Insights

Insight 1: We need to keep track of the maximum value encountered so far along the path from the root to the current node.
Insight 2: Depth-First Search (DFS) is a natural fit for traversing the tree and maintaining the path information.
Insight 3: We can use a helper function that takes the current node and the maximum value seen so far as input.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

This problem involves: Tree, Depth-First Search, Breadth-First Search, Binary Tree.

Companies

Asked at: Docusign, josh technology.

Count Good Nodes in Binary Tree | LeetCode 1448

Medium

Tree Depth-First Search Breadth-First Search Binary Tree

Time: O(n)Space: O(h)

Solution Code

Available in:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int goodNodes(TreeNode root) {
        return dfs(root, -10001);
    }

    private int dfs(TreeNode node, int max_val) {
        if (node == null) {
            return 0;
        }

        int res = node.val >= max_val ? 1 : 0;
        max_val = Math.max(max_val, node.val);
        res += dfs(node.left, max_val);
        res += dfs(node.right, max_val);
        return res;
    }
}

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Problem Info

Description

Problem 1448: Count Good Nodes in Binary Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Binary Tree

Companies

Docusignjosh technology

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1448

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize a counter to 0 to store the number of good nodes.
2
Step 2: Define a recursive helper function `dfs(node, max_val)` where `node` is the current node being visited and `max_val` is the maximum value seen from root to `node`.
3
Step 3: Base case for the recursion: If the current `node` is `None`, return 0 (no good nodes found in an empty subtree).
4
Step 4: Check if the current `node.val` is greater than or equal to `max_val`. If it is, increment the counter by 1 (it's a good node).
5
Step 5: Update `max_val` to be the maximum of the current `max_val` and `node.val`.
6
Step 6: Recursively call `dfs(node.left, max_val)` and add the result to the counter.
7
Step 7: Recursively call `dfs(node.right, max_val)` and add the result to the counter.
8
Step 8: Return the final count.
9
Step 9: Call the `dfs` function starting from the `root` node with an initial `max_val` smaller than any possible node value (e.g., -10001).
10
Step 10: Return the value returned by the `dfs` function as the result.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(h)

📝

Why?

The time complexity is O(n) because each node is visited exactly once. The space complexity depends on the height of the tree due to recursive calls. In the worst-case scenario (skewed tree), the height is n, and in the best-case scenario (balanced tree), the height is log n.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Breadth-First Search (BFS)

Instead of DFS, we can use BFS. We'd need to keep track of the maximum value seen so far for each path explored. We could store this information along with the node in the queue.

Time: O(n)Space: O(w), where w is the maximum width of the tree. In the worst case (a complete binary tree), w can be O(n).

Trade-offs: BFS can be more memory-intensive than DFS if the tree is very wide. DFS is generally preferred due to its lower space complexity (O(h)) compared to BFS's O(w), where h is tree height and w is tree width, especially when the tree's width is significantly larger than its height.

✨

Key Insights

1Insight 1: We need to keep track of the maximum value encountered so far along the path from the root to the current node.
2Insight 2: Depth-First Search (DFS) is a natural fit for traversing the tree and maintaining the path information.
3Insight 3: We can use a helper function that takes the current node and the maximum value seen so far as input.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is None). The code correctly handles this case by returning 0 in the base case of the DFS function.
Edge case 2: A tree with only one node (the root). The root node is always a good node and the code correctly identifies it and returns 1.
Edge case 3: All nodes have the same value. The code will correctly count all nodes as good since each node will be >= max_val along the path.
Edge case 4: Nodes with negative values. Using a sufficiently small initial value for 'max_val' (e.g., -10001) will handle negative node values correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not initializing the maximum value correctly. If the initial maximum value is not small enough (less than all possible node values), incorrect good nodes might be counted.

✗Common mistake 2: Not updating the maximum value correctly after visiting a node. Failing to update max_val with the current node's value when appropriate leads to incorrect results.

✗Common mistake 3: Forgetting the base case in the recursive DFS function (handling the null node). This will cause a runtime error.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing data streams to identify records that represent new peaks or best values so far. This can be used in stock market analysis or weather monitoring.

🏦Real-world scenario 2: In game development, identifying the highest score achieved by a player during a game session. Each node could represent a level, and the value represents the score in that level. 'Good' nodes would represent levels where the player achieved their highest score up to that point.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Most tree problems can be solved recursively. The base case is usually a null node. Think about what information needs to flow up vs down the tree.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Not handling null nodes properly. Always check if the current node exists before accessing its children.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

How do I choose between recursive and iterative tree traversal?

Use recursive traversal for cleaner code when stack depth isn't a concern (tree height < ~1000). Use iterative with explicit stack for very deep trees to avoid stack overflow. Iterative is also preferred when you need fine-grained control over the traversal process.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count Good Nodes in Binary Tree"?

For this medium-level Tree problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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