What is the time complexity of Find Bottom Left Tree Value?

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

What data structures are used to solve Find Bottom Left Tree Value?

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 Find Bottom Left Tree Value?

Insight 1: Breadth-First Search (BFS) is well-suited for traversing a tree level by level, making it easy to identify the last row. Insight 2: By processing the right child before the left child in the BFS, the last node processed in the queue will always be the leftmost node of the last level.

Which companies ask LeetCode 513. Find Bottom Left Tree Value in interviews?

This problem is frequently asked at josh technology.

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Find Bottom Left Tree Value - LeetCode 513 Solution

Q: How to solve Find Bottom Left Tree Value?

The solution utilizes Breadth-First Search (BFS) to traverse the binary tree level by level. The key idea is to add the right child to the queue *before* adding the left child. This ensures that when processing each level, we are moving from right to left. As a result, the last node we encounter (and the last one left in the queue) will be the leftmost node in the bottommost row of the tree. We keep updating `current_node` variable for each node visited by the BFS algorithm. Finally, the `current_node.val` stores the value of the leftmost node in the last row.

Find Bottom Left Tree Value - Complete Solution Guide

Find Bottom Left Tree Value is LeetCode problem 513, 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 the root of a binary tree, return the leftmost value in the last row of the tree. Example 1: Input: root = [2,1,3] Output: 1 Example 2: Input: root = [1,2,3,4,null,5,6,null,null,7] Output: 7 Constraints: The number of nodes in the tree is in the range [1, 10 4 ] . -2 31 <= Node.val <= 2 31 - 1

Detailed Explanation

The problem asks us to find the value of the leftmost node in the last (deepest) row of a given binary tree. The input is the root node of the tree, and the output is a single integer representing the value of the leftmost node in the bottom row.

Solution Approach

The solution utilizes Breadth-First Search (BFS) to traverse the binary tree level by level. The key idea is to add the right child to the queue *before* adding the left child. This ensures that when processing each level, we are moving from right to left. As a result, the last node we encounter (and the last one left in the queue) will be the leftmost node in the bottommost row of the tree. We keep updating `current_node` variable for each node visited by the BFS algorithm. Finally, the `current_node.val` stores the value of the leftmost node in the last row.

Step-by-Step Algorithm

Step 1: Initialize a queue with the root node.
Step 2: Initialize a variable `current_node` to store the last visited node; initially, it's set to the root.
Step 3: While the queue is not empty, dequeue a node from the queue and update the `current_node` to the value of the dequeued node.
Step 4: Enqueue the right child of the dequeued node (if it exists).
Step 5: Enqueue the left child of the dequeued node (if it exists).
Step 6: Repeat steps 3-5 until the queue is empty.
Step 7: Return the value of the `current_node` because after all the nodes are visted it contains the last visited node which is the left most node in last level

Key Insights

Insight 1: Breadth-First Search (BFS) is well-suited for traversing a tree level by level, making it easy to identify the last row.
Insight 2: By processing the right child before the left child in the BFS, the last node processed in the queue will always be the leftmost node of the last level.
Insight 3: We don't need to explicitly track the level depth. Since BFS processes level by level, the last node we dequeue is guaranteed to be the leftmost node in the last row.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(w)

Topics

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

Companies

Asked at: josh technology.

Find Bottom Left Tree Value | LeetCode 513

Medium

Tree Depth-First Search Breadth-First Search Binary Tree

Time: O(n)Space: O(w)

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.LinkedList;
import java.util.Queue;

class Solution {
    public int findBottomLeftValue(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        
        TreeNode current_node = root;
        while (!queue.isEmpty()) {
            current_node = queue.poll();
            
            if (current_node.right != null) {
                queue.add(current_node.right);
            }
            
            if (current_node.left != null) {
                queue.add(current_node.left);
            }
        }
        
        return current_node.val;
    }
}

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Problem Info

Description

Problem 513: Find Bottom Left Tree Value - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Binary Tree

Companies

josh technology

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:513

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: Initialize a queue with the root node.
2
Step 2: Initialize a variable `current_node` to store the last visited node; initially, it's set to the root.
3
Step 3: While the queue is not empty, dequeue a node from the queue and update the `current_node` to the value of the dequeued node.
4
Step 4: Enqueue the right child of the dequeued node (if it exists).
5
Step 5: Enqueue the left child of the dequeued node (if it exists).
6
Step 6: Repeat steps 3-5 until the queue is empty.
7
Step 7: Return the value of the `current_node` because after all the nodes are visted it contains the last visited node which is the left most node in last level

💡

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(w)

📝

Why?

The time complexity is O(n) because we visit each node of the tree once during the BFS traversal. The space complexity is O(w) because the queue will hold at most all the nodes at the widest level of the tree. In the worst-case scenario (a complete binary tree), the widest level is approximately half the total number of nodes (n/2), leading to O(n) space complexity, but a balanced tree will have a smaller width than the total number of nodes.

📖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

Depth-First Search (DFS)

Perform a DFS traversal, keeping track of the maximum depth reached so far. Update the leftmost value whenever a node at a greater depth is encountered.

Time: O(n)Space: O(h), where h is the height of the tree. In the worst case (skewed tree), h can be n, so it can be O(n).

Trade-offs: DFS might be slightly more complex to implement correctly, as it requires managing the current depth and updating the leftmost value accordingly. The space complexity depends on the height of the tree which can be larger than the widest level of the tree.

✨

Key Insights

1Insight 1: Breadth-First Search (BFS) is well-suited for traversing a tree level by level, making it easy to identify the last row.
2Insight 2: By processing the right child before the left child in the BFS, the last node processed in the queue will always be the leftmost node of the last level.
3Insight 3: We don't need to explicitly track the level depth. Since BFS processes level by level, the last node we dequeue is guaranteed to be the leftmost node in the last row.

📚

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 by checking if the root is None and then there is nothing to add to the queue and return the last element in queue, which is none.
Edge case 2: Single-node tree. The root is the bottom left value. The algorithm initializes the queue with the root and then processes it. Since there are no children, the loop terminates and the root's value is returned which is correct.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the tree is empty or contains only one node.

✗Common mistake 2: Enqueuing the left child before the right child in BFS. This would cause the rightmost node of the last level to be returned instead of the leftmost.

✗Common mistake 3: Not understanding the level-by-level traversal of BFS and trying to use DFS without correctly keeping track of the depth.

✅

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: Determining the first element to display in a deeply nested hierarchical structure, such as a file system directory tree.

🏦Real-world scenario 2: Finding the earliest task to execute in a dependency graph where tasks are arranged in a tree-like structure. In game AI, useful for determining the most important action to take in a decision tree.

💡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: November 1, 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.

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 "Find Bottom Left Tree Value"?

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?