What is the time complexity of Distribute Coins in Binary Tree?

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

What data structures are used to solve Distribute Coins in Binary Tree?

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 Distribute Coins in Binary Tree?

Insight 1: The core idea is to realize that the number of moves is determined by the 'balance' of each node. The balance is defined as the difference between the number of coins a node has and the number of coins it *should* have (which is 1). Insight 2: We can use Depth-First Search (DFS) to traverse the tree and calculate the balance for each node in a bottom-up manner. The absolute value of the balance represents the number of moves needed involving that node.

Which companies ask LeetCode 979. Distribute Coins in Binary Tree in interviews?

This problem is frequently asked at PhonePe.

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Distribute Coins in Binary Tree - LeetCode 979 Solution

Distribute Coins in Binary Tree - Complete Solution Guide

Distribute Coins in Binary Tree is LeetCode problem 979, 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 with n nodes where each node in the tree has node.val coins. There are n coins in total throughout the whole tree. In one move, we may choose two adjacent nodes and move one coin from one node to another. A move may be from parent to child, or from child to parent. Return the minimum number of moves required to make every node have exactly one coin . Example 1: Input: root = [3,0,0] Output: 2 Explanation: From the root of the tree, we move one coin to its

Detailed Explanation

The problem asks us to find the minimum number of moves required to distribute coins in a binary tree such that each node has exactly one coin. A move consists of transferring a coin between adjacent nodes (parent to child or child to parent). The total number of coins in the tree is equal to the number of nodes.

Solution Approach

The solution uses a Depth-First Search (DFS) algorithm to traverse the binary tree. For each node, we calculate its balance by recursively calculating the balance of its left and right children. The absolute values of the left and right children balances are added to a global `moves` variable. Then, the balance of the current node is calculated as the sum of its own value, the left balance, the right balance, minus 1 (since each node should ideally have one coin). This balance is returned to the parent node to be used in the parent's balance calculation.

Step-by-Step Algorithm

Step 1: Initialize a global variable `moves` to 0. This variable will store the total number of moves required.
Step 2: Define a recursive DFS function that takes a node as input.
Step 3: In the DFS function, if the node is null, return 0 (no balance).
Step 4: Recursively call the DFS function on the left child and store the returned balance in `left_balance`.
Step 5: Recursively call the DFS function on the right child and store the returned balance in `right_balance`.
Step 6: Add the absolute values of `left_balance` and `right_balance` to the `moves` variable.
Step 7: Calculate the current node's balance as `node.val + left_balance + right_balance - 1`. This represents the number of coins this node is either over or under.
Step 8: Return the current node's balance to the parent node.
Step 9: Call the DFS function on the root node.
Step 10: Return the `moves` variable, which now contains the total number of moves required.

Key Insights

Insight 1: The core idea is to realize that the number of moves is determined by the 'balance' of each node. The balance is defined as the difference between the number of coins a node has and the number of coins it *should* have (which is 1).
Insight 2: We can use Depth-First Search (DFS) to traverse the tree and calculate the balance for each node in a bottom-up manner. The absolute value of the balance represents the number of moves needed involving that node.
Insight 3: The balance can be positive (excess coins) or negative (deficit of coins). The excess/deficit can be pushed to/pulled from the parent node.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

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

Companies

Asked at: PhonePe.

Distribute Coins in Binary Tree | LeetCode 979

Medium

Tree Depth-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 {
    int moves = 0;

    public int distributeCoins(TreeNode root) {
        moves = 0;
        dfs(root);
        return moves;
    }

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

        int left_balance = dfs(node.left);
        int right_balance = dfs(node.right);

        moves += Math.abs(left_balance) + Math.abs(right_balance);

        return node.val + left_balance + right_balance - 1;
    }
}

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Problem Info

Description

Problem 979: Distribute Coins in Binary Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

PhonePe

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:979

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 variable `moves` to 0. This variable will store the total number of moves required.
2
Step 2: Define a recursive DFS function that takes a node as input.
3
Step 3: In the DFS function, if the node is null, return 0 (no balance).
4
Step 4: Recursively call the DFS function on the left child and store the returned balance in `left_balance`.
5
Step 5: Recursively call the DFS function on the right child and store the returned balance in `right_balance`.
6
Step 6: Add the absolute values of `left_balance` and `right_balance` to the `moves` variable.
7
Step 7: Calculate the current node's balance as `node.val + left_balance + right_balance - 1`. This represents the number of coins this node is either over or under.
8
Step 8: Return the current node's balance to the parent node.
9
Step 9: Call the DFS function on the root node.
10
Step 10: Return the `moves` variable, which now contains the total number of moves required.

💡

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 we visit each of the 'n' nodes in the tree exactly once. The space complexity is determined by the depth of the recursion, which is at most the height of the tree. This recursion depth dictates the size of the call stack, which is the main contributor to space usage.

📖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

Iterative DFS with Stack

An iterative DFS approach using a stack to avoid recursion. The logic for balance calculation remains the same, but the tree traversal is handled iteratively.

Time: O(n)Space: O(n) in the worst case (skewed tree) to store nodes in the stack, and O(log n) on average(balanced tree).

Trade-offs: Iterative approach avoids the potential stack overflow issues that might occur with very deep trees in a recursive approach, at the cost of a slightly more complex implementation and potentially more space usage in some cases.

✨

Key Insights

1Insight 1: The core idea is to realize that the number of moves is determined by the 'balance' of each node. The balance is defined as the difference between the number of coins a node has and the number of coins it *should* have (which is 1).
2Insight 2: We can use Depth-First Search (DFS) to traverse the tree and calculate the balance for each node in a bottom-up manner. The absolute value of the balance represents the number of moves needed involving that node.
3Insight 3: The balance can be positive (excess coins) or negative (deficit of coins). The excess/deficit can be pushed to/pulled from the parent node.

📚

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 handles this correctly because the base case of the DFS function returns 0 when a null node is encountered, preventing any errors.
Edge case 2: A single node tree. The algorithm still correctly calculates the moves as 0, as the balance would be node.val - 1, which becomes 0 if node.val is equal to 1 (as mandated by the problem constraints that sum of all Node.val = n = number of nodes). The abs(balance) added to move is then 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the absolute value of the balance when calculating moves. The number of moves is always positive, regardless of whether a node is over or under the required number of coins.

✗Common mistake 2: Forgetting to subtract 1 from the node's value when calculating the balance. Each node should have exactly one coin in the end state.

✅

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: Resource allocation in a network. This problem is similar to distributing resources (like bandwidth or memory) across a network of interconnected nodes to meet certain requirements, minimizing transfer costs.

🏦Real-world scenario 2: Inventory management in a supply chain. Optimizing the flow of goods in a supply chain to ensure that each distribution center has the right amount of inventory can be modeled using similar techniques to minimize transportation costs.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: August 19, 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

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

@TechLeadJul 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

Frequently Asked Questions

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Distribute Coins 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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