What is the time complexity of Number of Good Leaf Nodes Pairs?

The optimal solution has a time complexity of O(n^2).

What data structures are used to solve Number of Good Leaf Nodes Pairs?

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

What are the key insights for solving Number of Good Leaf Nodes Pairs?

Insight 1: We need to use Depth-First Search (DFS) to traverse the tree and find all leaf nodes. Insight 2: For each node, we need to keep track of the distances from its descendant leaf nodes to itself.

Which companies ask LeetCode 1530. Number of Good Leaf Nodes Pairs in interviews?

This problem is frequently asked at ByteDance, josh technology.

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Number of Good Leaf Nodes Pairs - LeetCode 1530 Solution

Number of Good Leaf Nodes Pairs - Complete Solution Guide

Number of Good Leaf Nodes Pairs is LeetCode problem 1530, 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

You are given the root of a binary tree and an integer distance . A pair of two different leaf nodes of a binary tree is said to be good if the length of the shortest path between them is less than or equal to distance . Return the number of good leaf node pairs in the tree. Example 1: Input: root = [1,2,3,null,4], distance = 3 Output: 1 Explanation: The leaf nodes of the tree are 3 and 4 and the length of the shortest path between them is 3. This is the only good pair. Example 2: Input: root =

Detailed Explanation

The problem asks us to find the number of 'good' leaf node pairs in a binary tree. A 'good' pair is defined as two distinct leaf nodes whose shortest path length is less than or equal to a given 'distance'. The input is the root of the binary tree and the integer 'distance'. The output is the count of good leaf node pairs.

Solution Approach

The solution employs a recursive Depth-First Search (DFS) approach. The `dfs` function returns a list of distances from leaf nodes in the subtree rooted at the current node to the current node. For each node, it recursively calls `dfs` on its left and right children. If the current node is a leaf node, it returns a list containing only 0 (since the distance from a leaf to itself is 0). If the current node is not a leaf node, it gets the lists of distances from its left and right children. It then increments each distance in these lists (to account for the edge connecting the child to the current node). Finally, it iterates through all possible pairs of distances (one from the left subtree and one from the right subtree) and increments the global `count` if their sum is less than or equal to `distance`. The function returns a list of distances from descendant leaf nodes to the current node, excluding distances which are already greater than or equal to 'distance'.

Step-by-Step Algorithm

Step 1: Initialize a global counter `count` to 0. This counter will store the number of good leaf node pairs.
Step 2: Define a recursive function `dfs(node, distance)` that takes a node and the maximum allowed distance as input.
Step 3: Base Case: If the node is null, return an empty list. If the node is a leaf node (no left and right children), return a list containing only 0 (distance from itself).
Step 4: Recursively call `dfs` on the left and right children of the current node to obtain lists of distances from leaf nodes in those subtrees to their respective children.
Step 5: Increment each distance in the lists returned from the left and right children by 1 to represent the distance from the leaf nodes to the current node.
Step 6: Iterate through all possible pairs of distances (one from the left subtree and one from the right subtree). If the sum of any pair of distances is less than or equal to the given `distance`, increment the global `count`.
Step 7: Combine the distances from the left and right subtrees into a single list. Filter this list to keep only distances that are strictly less than the given `distance`.
Step 8: Return the filtered list of distances from the current node to its descendant leaf nodes.
Step 9: Call the `dfs` function with the root of the tree and the given `distance`.
Step 10: Return the final value of the `count`.

Key Insights

Insight 1: We need to use Depth-First Search (DFS) to traverse the tree and find all leaf nodes.
Insight 2: For each node, we need to keep track of the distances from its descendant leaf nodes to itself.
Insight 3: When we reach a node, we can combine the distances from leaf nodes in its left and right subtrees to identify good pairs.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n)

Topics

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

Companies

Asked at: ByteDance, josh technology.

Number of Good Leaf Nodes Pairs | LeetCode 1530

Medium

Tree Depth-First Search Binary Tree

Time: O(n^2)Space: O(n)

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;
 *     }
 * }
 */
import java.util.ArrayList;
import java.util.List;

class Solution {
    private int count = 0;

    public int countPairs(TreeNode root, int distance) {
        this.count = 0;
        dfs(root, distance);
        return this.count;
    }

    private List<Integer> dfs(TreeNode node, int distance) {
        if (node == null) {
            return new ArrayList<>();
        }

        if (node.left == null && node.right == null) {
            List<Integer> list = new ArrayList<>();
            list.add(0);
            return list;
        }

        List<Integer> left_dists_from_child = dfs(node.left, distance);
        List<Integer> right_dists_from_child = dfs(node.right, distance);

        List<Integer> left_dists_to_node = new ArrayList<>();
        for (int d : left_dists_from_child) {
            left_dists_to_node.add(d + 1);
        }

        List<Integer> right_dists_to_node = new ArrayList<>();
        for (int d : right_dists_from_child) {
            right_dists_to_node.add(d + 1);
        }

        for (int d1 : left_dists_to_node) {
            for (int d2 : right_dists_to_node) {
                if (d1 + d2 <= distance) {
                    this.count += 1;
                }
            }
        }

        List<Integer> all_dists_to_node = new ArrayList<>();
        all_dists_to_node.addAll(left_dists_to_node);
        all_dists_to_node.addAll(right_dists_to_node);
        
        List<Integer> result = new ArrayList<>();
        for (int d : all_dists_to_node) {
            if (d < distance) {
                result.add(d);
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 1530: Number of Good Leaf Nodes Pairs - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

ByteDancejosh technology

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1530

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 global counter `count` to 0. This counter will store the number of good leaf node pairs.
2
Step 2: Define a recursive function `dfs(node, distance)` that takes a node and the maximum allowed distance as input.
3
Step 3: Base Case: If the node is null, return an empty list. If the node is a leaf node (no left and right children), return a list containing only 0 (distance from itself).
4
Step 4: Recursively call `dfs` on the left and right children of the current node to obtain lists of distances from leaf nodes in those subtrees to their respective children.
5
Step 5: Increment each distance in the lists returned from the left and right children by 1 to represent the distance from the leaf nodes to the current node.
6
Step 6: Iterate through all possible pairs of distances (one from the left subtree and one from the right subtree). If the sum of any pair of distances is less than or equal to the given `distance`, increment the global `count`.
7
Step 7: Combine the distances from the left and right subtrees into a single list. Filter this list to keep only distances that are strictly less than the given `distance`.
8
Step 8: Return the filtered list of distances from the current node to its descendant leaf nodes.
9
Step 9: Call the `dfs` function with the root of the tree and the given `distance`.
10
Step 10: Return the final value of the `count`.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the nested loops that compare distances from left and right subtrees. The space complexity is determined by the maximum depth of the recursion stack and the space required to store distance lists at each node in the call stack.

📖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

Precompute all pairs shortest paths

Find all leaf nodes. Then compute all pairs shortest paths using a graph traversal algorithm like BFS on the tree structure. Then count the pairs with distance <= given distance.

Time: O(N^2), where N is number of nodes in tree in the worst case.Space: O(N^2) to store the all pairs shortest path matrix in the worst case.

Trade-offs: This approach is generally less efficient because of the larger memory usage and potentially higher constant factors. However, if we need to answer multiple queries for different distances on the same tree, precomputing all pairs shortest paths might become beneficial.

✨

Key Insights

1Insight 1: We need to use Depth-First Search (DFS) to traverse the tree and find all leaf nodes.
2Insight 2: For each node, we need to keep track of the distances from its descendant leaf nodes to itself.
3Insight 3: When we reach a node, we can combine the distances from leaf nodes in its left and right subtrees to identify good pairs.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree: The solution correctly handles an empty tree because the initial check `if not node:` in the `dfs` function returns an empty list, preventing any further processing.
Edge case 2: Tree with only one node (leaf node): The solution correctly handles a tree with only one node because the second check `if not node.left and not node.right:` in the `dfs` function returns a list containing only 0. This prevents any further processing and does not affect the count.
Edge case 3: distance = 1 and deep skewed tree: The algorithm might take longer to process since leaf nodes can be far from root.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to increment the distances after the recursive calls on left and right children. Without incrementing, the distances would represent the distances from the leaf nodes to the *children*, not to the current node.

✗Common mistake 2: Not filtering the distances before returning. This can lead to unnecessary comparisons and increase the time complexity.

✅

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: Network analysis: Identifying pairs of computers in a network that are within a certain number of hops of each other.

🏦Real-world scenario 2: Social network analysis: Finding pairs of users in a social network who are connected by a path of a certain length.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: August 30, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

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 "Number of Good Leaf Nodes Pairs"?

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.

🤖 Need help with this problem?