LeetCode 515: Find Largest Value in Each Tree Row Solution in Python – A Step-by-Step Guide

What Is LeetCode 515: Find Largest Value in Each Tree Row?

In LeetCode 515: Find Largest Value in Each Tree Row, you’re given the root of a binary tree, and your task is to return a list of the largest values in each level (row) from top to bottom. For example, in a tree with root 1, left child 3, and right child 2, the result is [1, 3]—1 for level 1, 3 for level 2 (since 3 > 2). This problem tests your ability to process a tree by levels, building on LeetCode 102: Binary Tree Level Order Traversal.

Problem Statement

• Input: Root node of a binary tree (TreeNode with val, left, right).
• Output: List of integers—largest value in each level, top to bottom.
• Rules: Tree is non-empty; levels start at root (level 1).

Constraints

• Number of nodes: 1 to 10⁴.
• Node values: -2³¹ to 2³¹ - 1.

Examples

• Input: [1,3,2,5,3,null,9]
• Output: [1,3,9]
• Tree:
``` 1 / \ 3 2 / \ \ 5 3 9 ```
• Level 1: [1] → 1.
• Level 2: [3, 2] → 3.
• Level 3: [5, 3, 9] → 9.
• Input: [1,2,3]
• Output: [1,3]
• Tree:
``` 1 / \ 2 3 ```
• Level 1: [1] → 1.
• Level 2: [2, 3] → 3.
• Input: [1]
• Output: [1]
• Single node: [1] → 1.

Understanding the Problem: Finding the Biggest in Each Row

To solve LeetCode 515: Find Largest Value in Each Tree Row in Python, we need a method to traverse the tree, group nodes by level, and find the maximum value in each group. A naive approach might traverse randomly and sort later, but with up to 10⁴ nodes, we need an organized strategy. We’ll explore:

• Best Solution (BFS with Level Tracking): O(n) time, O(w) space (w = max width)—level-based and efficient.
• Alternative Solution (DFS with Depth Mapping): O(n) time, O(h) space (h = height)—recursive and compact.

Let’s dive into the BFS solution with a friendly breakdown!

Best Solution: BFS with Level Tracking

Why BFS Wins

The BFS with level tracking is the best for LeetCode 515 because it naturally processes the tree level by level, making it easy to find the maximum value in each row in a single pass. It runs in O(n) time (n nodes) and O(w) space (w is max width), offering a clear, intuitive approach. It’s like sweeping through the tree floor by floor, spotlighting the biggest value each time!

How It Works

Think of this as a level-by-level treasure sweep:

• Step 1: Use a Queue:
• Start with the root in a queue.
• Step 2: Process Each Level:
• For each level, track the size, dequeue all nodes, and find the max value.
• Enqueue children for the next level.
• Step 3: Build Result:
• Append each level’s max to the result list.
• Why It Works:
• BFS ensures level order.
• One pass gets all maxes.

It’s like a max-value level scanner!

Step-by-Step Example

Example: [1,3,2,5,3,null,9]

• Tree:
``` 1 / \ 3 2 / \ \ 5 3 9 ```
• Init: Queue = [1], result = [].
• Level 1:
• Size = 1, dequeue 1, max = 1.
• Enqueue 3, 2.
• Queue = [3, 2], result = [1].
• Level 2:
• Size = 2, dequeue 3, 2, max = max(3, 2) = 3.
• Enqueue 5, 3, 9.
• Queue = [5, 3, 9], result = [1, 3].
• Level 3:
• Size = 3, dequeue 5, 3, 9, max = max(5, 3, 9) = 9.
• Queue = [], result = [1, 3, 9].
• Result: [1, 3, 9].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codefrom collections import deque

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def largestValues(self, root: TreeNode) -> List[int]:
        if not root:
            return []

        # Step 1: Initialize queue and result
        queue = deque([root])
        result = []

        # Step 2: BFS traversal
        while queue:
            level_size = len(queue)
            level_max = float('-inf')  # Track max in level

            # Process current level
            for _ in range(level_size):
                node = queue.popleft()
                level_max = max(level_max, node.val)

                # Enqueue children
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)

            # Add max to result
            result.append(level_max)

        # Step 3: Return list of maxes
        return result

• Lines 11-12: Handle empty tree.
• Lines 15-16: Queue with root, empty result list.
• Line 19: Loop until queue is empty.
• Lines 20-22: Size and max for current level.
• Lines 25-30: Dequeue node, update max, enqueue kids.
• Line 33: Append level’s max to result.
• Time Complexity: O(n)—visit each node once.
• Space Complexity: O(w)—queue holds max width.

It’s like a level-by-level max finder!

Alternative Solution: DFS with Depth Mapping

Why an Alternative Approach?

The DFS with depth mapping uses recursion to explore the tree, mapping each node’s value to its depth, then finding the max per level. It’s O(n) time and O(h) space (h = height)—compact and elegant but requires post-processing. Great for recursion lovers!

How It Works

Picture this as a depth-first treasure map:

• Step 1: Recursively traverse, tracking depth.
• Step 2: Store values in a depth-to-values dictionary.
• Step 3: Extract max from each depth.

It’s like mapping the tree’s treasures by floor!

Step-by-Step Example

Example: [1,2,3]

• Tree:
``` 1 / \ 2 3 ```
• Init: depth_map = {}.
• DFS:
• Root (1): Depth 0, depth_map[0] = [1].
• Left (2): Depth 1, depth_map[1] = [2].
• Right (3): Depth 1, depth_map[1] = [2, 3].
• Result: [max([1]), max([2, 3])] = [1, 3].

Code for DFS Approach

Copy codeclass Solution:
    def largestValues(self, root: TreeNode) -> List[int]:
        if not root:
            return []

        # Step 1: Map depth to values
        depth_map = {}

        def dfs(node, depth):
            if not node:
                return

            # Add value to depth’s list
            if depth not in depth_map:
                depth_map[depth] = []
            depth_map[depth].append(node.val)

            # Recurse
            dfs(node.left, depth + 1)
            dfs(node.right, depth + 1)

        # Step 2: Run DFS
        dfs(root, 0)

        # Step 3: Get max per level
        return [max(values) for values in depth_map.values()]

• Lines 6-7: Dictionary for depth mapping.
• Lines 10-18: DFS:
• Base case: null.
• Append value to depth’s list.
• Recurse left and right.
• Line 22: Build result with max per depth.
• Time Complexity: O(n)—visit each node.
• Space Complexity: O(h)—recursion stack (map size is output).

It’s a depth-mapped max hunter!

Comparing the Two Solutions

• BFS (Best):
• Pros: O(n) time, O(w) space, level clarity.
• Cons: Queue overhead.
• DFS (Alternative):
• Pros: O(n) time, O(h) space, elegant.
• Cons: Post-processing.

BFS wins for intuition!

Additional Examples and Edge Cases

• ** [1]: [1].
• ** [1,null,2]: [1, 2].
• ** [1,2,null,3]: [1, 2, 3].

BFS handles them all!

Complexity Recap

• BFS: Time O(n), Space O(w).
• DFS: Time O(n), Space O(h).

BFS is the level-sweeping star!

Key Takeaways

• BFS: Level order shines—learn at Python Basics!
• DFS: Depth with mapping.
• Trees: Max values are fun.
• Python: Queues or dicts, your choice!

Final Thoughts: Hunt Those Maxes!

LeetCode 515: Find Largest Value in Each Tree Row in Python is a delightful tree challenge. BFS with level tracking is your clear path, while DFS adds recursive flair. Want more? Try LeetCode 102: Binary Tree Level Order Traversal or LeetCode 513: Find Bottom Left Tree Value. Ready to traverse? Head to Solve LeetCode 515 on LeetCode and find those biggest values today!