What is the time complexity of Sum Root to Leaf Numbers?

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

What data structures are used to solve Sum Root to Leaf Numbers?

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 Sum Root to Leaf Numbers?

Insight 1: Depth-First Search (DFS) is a natural fit for traversing the tree and exploring each root-to-leaf path. Insight 2: We can pass the 'current number' being formed down the tree in the recursive calls, avoiding string manipulations and keeping the code efficient.

Which companies ask LeetCode 129. Sum Root to Leaf Numbers in interviews?

This problem is frequently asked at ServiceNow, Visa.

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Sum Root to Leaf Numbers - LeetCode 129 Solution

Sum Root to Leaf Numbers - Complete Solution Guide

Sum Root to Leaf Numbers is LeetCode problem 129, 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 containing digits from 0 to 9 only. Each root-to-leaf path in the tree represents a number. For example, the root-to-leaf path 1 -> 2 -> 3 represents the number 123 . Return the total sum of all root-to-leaf numbers . Test cases are generated so that the answer will fit in a 32-bit integer. A leaf node is a node with no children. Example 1: Input: root = [1,2,3] Output: 25 Explanation: The root-to-leaf path 1->2 represents the number 12 . The root-to-leaf

Detailed Explanation

The problem asks us to calculate the sum of all numbers formed by the paths from the root to each leaf node in a binary tree. Each node in the tree contains a digit from 0 to 9. A path from the root to a leaf represents a number where the digits are concatenated in the order they appear along the path. The goal is to traverse the tree, construct these numbers, and return their sum. For example, if a path is 1 -> 2 -> 3, the number formed is 123. A leaf node is a node with no left or right children. The constraints specify that the tree will have between 1 and 1000 nodes, node values are between 0 and 9, and the depth of the tree will not exceed 10. The answer must fit in a 32-bit integer.

Solution Approach

The solution uses a recursive Depth-First Search (DFS) approach. The `dfs` function takes a node and the 'current number' as input. At each node, it updates the 'current number' by multiplying it by 10 and adding the node's value. If the node is a leaf node, the 'current number' is returned. Otherwise, the function recursively calls itself on the left and right children, passing the updated 'current number'. The results of the recursive calls are summed and returned. The initial call to `dfs` is made with the root node and an initial 'current number' of 0. This traversal effectively explores all root-to-leaf paths and accumulates their numerical values.

Step-by-Step Algorithm

Step 1: Start with the root node and an initial current number of 0.
Step 2: Multiply the current number by 10 and add the value of the current node to it. This forms the number represented by the path from the root to the current node.
Step 3: Check if the current node is a leaf node (no left and no right children).
Step 4: If it is a leaf node, return the current number. This indicates that a complete path from root to leaf has been found.
Step 5: If it is not a leaf node, recursively call the DFS function on the left and right children, passing the updated current number.
Step 6: Sum the results returned by the left and right recursive calls. This sums the numbers formed by all the paths passing through this node.
Step 7: Return the sum from Step 6. This value propagates back up the call stack until it reaches the initial call, which returns the final sum of all root-to-leaf numbers.

Key Insights

Insight 1: Depth-First Search (DFS) is a natural fit for traversing the tree and exploring each root-to-leaf path.
Insight 2: We can pass the 'current number' being formed down the tree in the recursive calls, avoiding string manipulations and keeping the code efficient.
Insight 3: Checking for a leaf node is essential to determine when to add the formed number to the total sum. A leaf node is where a path ends.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

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

Companies

Asked at: ServiceNow, Visa.

Sum Root to Leaf Numbers | LeetCode 129

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 {
    public int sumNumbers(TreeNode root) {
        return dfs(root, 0);
    }

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

        current_number = current_number * 10 + node.val;

        if (node.left == null && node.right == null) {
            return current_number;
        }

        int left_sum = dfs(node.left, current_number);
        int right_sum = dfs(node.right, current_number);

        return left_sum + right_sum;
    }
}

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Problem Info

Description

Problem 129: Sum Root to Leaf Numbers - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

ServiceNowVisa

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:129

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

7 steps

1
Step 1: Start with the root node and an initial current number of 0.
2
Step 2: Multiply the current number by 10 and add the value of the current node to it. This forms the number represented by the path from the root to the current node.
3
Step 3: Check if the current node is a leaf node (no left and no right children).
4
Step 4: If it is a leaf node, return the current number. This indicates that a complete path from root to leaf has been found.
5
Step 5: If it is not a leaf node, recursively call the DFS function on the left and right children, passing the updated current number.
6
Step 6: Sum the results returned by the left and right recursive calls. This sums the numbers formed by all the paths passing through this node.
7
Step 7: Return the sum from Step 6. This value propagates back up the call stack until it reaches the initial call, which returns the final sum of all root-to-leaf numbers.

💡

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 is O(H) due to the recursion stack. The depth of recursion can be at most the height of the tree. In the average case of a balanced tree, H is log(N), making the space complexity O(log(N)). However, in the worst case (skewed tree), H = N, resulting in O(N) space complexity.

📖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 Depth-First Search (DFS) with Stack

An alternative to recursion is to use an iterative DFS approach with a stack. This would involve explicitly managing a stack to keep track of the nodes to visit and the 'current number' associated with each node.

Time: O(N), where N is the number of nodes.Space: O(H) in average case, and O(N) in worst case (skewed tree), where H is the height of the tree.

Trade-offs: Iterative DFS avoids the overhead of recursion and may be slightly more efficient in some cases. It can also avoid stack overflow issues in very deep trees. However, the code may be more complex to write and understand.

Breadth-First Search (BFS)

BFS could be used but it would not be an optimal approach because it would visit nodes level by level rather than path by path. While it's possible, the code would be less readable and less efficient than DFS.

Time: O(N)Space: O(W), where W is the maximum width of the tree. Can be O(N) in worst case (complete binary tree)

Trade-offs: BFS is generally not preferred for this problem because DFS provides a more natural way to explore root-to-leaf paths.

✨

Key Insights

1Insight 1: Depth-First Search (DFS) is a natural fit for traversing the tree and exploring each root-to-leaf path.
2Insight 2: We can pass the 'current number' being formed down the tree in the recursive calls, avoiding string manipulations and keeping the code efficient.
3Insight 3: Checking for a leaf node is essential to determine when to add the formed number to the total sum. A leaf node is where a path ends.

📚

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 implicitly handles this because the initial check `if not node:` in the DFS function returns 0, which is the correct sum for an empty tree.
Edge case 2: A tree with only a root node (the root is also a leaf). The code correctly handles this because when the DFS reaches the root, it will be identified as a leaf and its value is returned as the 'current number'.
Edge case 3: All nodes have value 0. The algorithm will correctly calculate 0 as the sum of the paths. The multiplication and addition operations will keep the values as zero.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the tree is empty. Ensure the code correctly returns 0 when the root is None.

✗Common mistake 2: Forgetting to multiply the 'current number' by 10 before adding the current node's value. This leads to incorrect calculations.

✗Common mistake 3: Not properly identifying leaf nodes. The code must check for both left and right children being None to correctly determine if a node is a leaf.

✅

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: Pathfinding in a graph or network where each edge has a weight or cost associated with it. The problem can be adapted to find the total cost of all paths from a starting node to destination nodes.

🏦Real-world scenario 2: Decision tree traversal. Imagine a decision tree where each node represents a decision and each leaf represents an outcome with a numerical value. This algorithm could be used to calculate the sum of values for all possible outcomes.

💡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.

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: October 8, 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 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 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.

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 "Sum Root to Leaf Numbers"?

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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