What is the time complexity of Binary Tree Tilt?

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

What data structures are used to solve Binary Tree Tilt?

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

What are the key insights for solving Binary Tree Tilt?

Insight 1: The tilt of a node depends on the sum of the values of its left and right subtrees. Therefore, a post-order traversal is suitable because we need to calculate the subtree sums before we can determine the tilt of a parent node. Insight 2: We can use Depth-First Search (DFS) to traverse the tree and compute the subtree sums and tilts recursively.

Which companies ask LeetCode 563. Binary Tree Tilt in interviews?

This problem is frequently asked at Indeed.

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Binary Tree Tilt - LeetCode 563 Solution

Binary Tree Tilt - Complete Solution Guide

Binary Tree Tilt is LeetCode problem 563, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given the root of a binary tree, return the sum of every tree node's tilt . The tilt of a tree node is the absolute difference between the sum of all left subtree node values and all right subtree node values . If a node does not have a left child, then the sum of the left subtree node values is treated as 0 . The rule is similar if the node does not have a right child. Example 1: Input: root = [1,2,3] Output: 1 Explanation: Tilt of node 2 : |0-0| = 0 (no children) Tilt of node 3 : |0-0| = 0 (no

Detailed Explanation

The problem asks us to calculate the sum of the 'tilt' of each node in a given binary tree. The 'tilt' of a node is defined as the absolute difference between the sum of all node values in its left subtree and the sum of all node values in its right subtree. If a node doesn't have a left or right child, we treat the respective subtree sum as 0. The input is the root of the binary tree, and the output is the total tilt of the entire tree.

Solution Approach

The solution uses a recursive Depth-First Search (DFS) algorithm to traverse the binary tree. The DFS function computes the sum of the subtree rooted at each node and simultaneously calculates the tilt of that node. A global variable (or a class member in object-oriented languages) accumulates the total tilt of the entire tree. The DFS function returns the sum of the subtree rooted at the current node to be used by the parent node in its tilt calculation.

Step-by-Step Algorithm

Step 1: Initialize a variable `total_tilt` (or `tilt` in Java/C++) to 0 to store the sum of tilts.
Step 2: Define a recursive function `dfs(node)` that takes a node as input.
Step 3: Base case for the recursion: If the `node` is `None` (or `null`), return 0.
Step 4: Recursively call `dfs` on the left child: `left_sum = dfs(node.left)`.
Step 5: Recursively call `dfs` on the right child: `right_sum = dfs(node.right)`.
Step 6: Calculate the tilt of the current `node`: `tilt = abs(left_sum - right_sum)`.
Step 7: Add the calculated `tilt` to the `total_tilt` variable: `total_tilt += tilt`.
Step 8: Return the sum of the subtree rooted at the current `node`: `return node.val + left_sum + right_sum`.
Step 9: Call the `dfs` function with the `root` node.
Step 10: Return the `total_tilt`.

Key Insights

Insight 1: The tilt of a node depends on the sum of the values of its left and right subtrees. Therefore, a post-order traversal is suitable because we need to calculate the subtree sums before we can determine the tilt of a parent node.
Insight 2: We can use Depth-First Search (DFS) to traverse the tree and compute the subtree sums and tilts recursively.
Insight 3: The total tilt can be calculated incrementally during the DFS traversal.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

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

Companies

Asked at: Indeed.

Binary Tree Tilt | LeetCode 563

Easy

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 tilt = 0;
    public int findTilt(TreeNode root) {
        dfs(root);
        return tilt;
    }
    public int dfs(TreeNode node){
        if(node == null) return 0;
        int leftSum = dfs(node.left);
        int rightSum = dfs(node.right);
        tilt += Math.abs(leftSum - rightSum);
        return leftSum + rightSum + node.val;
    }
}

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Problem Info

Description

Problem 563: Binary Tree Tilt - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

Indeed

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:563

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 variable `total_tilt` (or `tilt` in Java/C++) to 0 to store the sum of tilts.
2
Step 2: Define a recursive function `dfs(node)` that takes a node as input.
3
Step 3: Base case for the recursion: If the `node` is `None` (or `null`), return 0.
4
Step 4: Recursively call `dfs` on the left child: `left_sum = dfs(node.left)`.
5
Step 5: Recursively call `dfs` on the right child: `right_sum = dfs(node.right)`.
6
Step 6: Calculate the tilt of the current `node`: `tilt = abs(left_sum - right_sum)`.
7
Step 7: Add the calculated `tilt` to the `total_tilt` variable: `total_tilt += tilt`.
8
Step 8: Return the sum of the subtree rooted at the current `node`: `return node.val + left_sum + right_sum`.
9
Step 9: Call the `dfs` function with the `root` node.
10
Step 10: Return the `total_tilt`.

💡

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 during the DFS traversal. The space complexity is determined by the maximum depth of the recursion stack, which is equal to the height of the tree. A balanced tree will have a height of log(n) whereas a skewed tree will have a height of 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

Iterative Approach using Stack

An iterative approach using a stack can be used to simulate the DFS traversal. This approach avoids recursion and can reduce the space complexity in certain cases.

Time: O(n)Space: O(n) in the worst case (skewed tree), O(log n) average case (balanced tree)

Trade-offs: While iterative solutions can be slightly more space-efficient in certain scenarios (especially in languages with tail-call optimization limitations), the recursive approach is generally more concise and easier to understand for this problem. The iterative approach would require explicit stack management.

✨

Key Insights

1Insight 1: The tilt of a node depends on the sum of the values of its left and right subtrees. Therefore, a post-order traversal is suitable because we need to calculate the subtree sums before we can determine the tilt of a parent node.
2Insight 2: We can use Depth-First Search (DFS) to traverse the tree and compute the subtree sums and tilts recursively.
3Insight 3: The total tilt can be calculated incrementally during the DFS traversal.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is None/null). The code handles this correctly because the base case of the DFS function returns 0 when the node is None, so the tilt will remain 0.
Edge case 2: A node with only one child (either left or right). The code handles this correctly because if a node doesn't have a left or right child, the respective subtree sum will be 0, as returned by the base case of the recursion. The tilt calculation will then simply be the absolute value of the existing child's subtree sum.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the base case (empty tree or null node) correctly. Failing to return 0 for null nodes will lead to incorrect subtree sums and therefore incorrect tilt calculations.

✗Common mistake 2: Not accumulating the total tilt correctly. It's crucial to add the calculated tilt of each node to the global `total_tilt` variable.

✗Common mistake 3: Calculating the tilt *after* the recursive calls, but before returning the subtree sum. This ensures we have the sums of the children to calculate the tilt of the parent node.

✅

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: Balancing of resources in a distributed system. The 'tilt' could represent the imbalance of resources between two branches or servers, and the overall 'tilt' could represent the total imbalance in the system. The algorithm can help identify areas where resources need to be reallocated.

🏦Real-world scenario 2: Network traffic analysis. In a network tree structure, the 'tilt' could represent the difference in traffic load between two sub-networks. The algorithm can help identify congested areas and optimize traffic routing.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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 24, 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

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

@DevExpert92Jun 29

❓

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Binary Tree Tilt"?

For this easy-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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