LeetCode 513: Find Bottom Left Tree Value Solution in Python – A Step-by-Step Guide

What Is LeetCode 513: Find Bottom Left Tree Value?

In LeetCode 513: Find Bottom Left Tree Value, you’re given the root of a binary tree, and your task is to find the value of the leftmost node in the last row (deepest level). For example, in a tree with root 2, left child 1, and right child 3, the bottom left value is 1 (only one level below root). This problem tests your ability to traverse a tree and track levels, building on concepts from LeetCode 104: Maximum Depth of Binary Tree.

Problem Statement

• Input: Root node of a binary tree (TreeNode with val, left, right).
• Output: Integer—value of the leftmost node in the bottommost level.
• Rules: Tree is non-empty; levels are counted from root (level 1).

Constraints

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

Examples

• Input: [2,1,3]
• Output: 1
• Tree:
``` 2 / \ 1 3 ```
• Bottom level: [1, 3], leftmost = 1.
• Input: [1,2,3,4,null,5,6,null,null,7]
• Output: 7
• Tree:
``` 1 / \ 2 3 / / \ 4 5 6 / 7 ```
• Bottom level: [7], leftmost = 7.
• Input: [1]
• Output: 1
• Single node, bottom left = 1.

Understanding the Problem: Digging to the Bottom Left

To solve LeetCode 513: Find Bottom Left Tree Value in Python, we need a method to traverse the tree, identify the deepest level, and pick the leftmost value there. A naive approach might traverse randomly, but with up to 10⁴ nodes, we need a systematic way to track levels and positions. We’ll explore:

• Best Solution (BFS Right-to-Left): O(n) time, O(w) space (w = max width)—intuitive and level-based.
• Alternative Solution (DFS with Depth): O(n) time, O(h) space (h = height)—clever but recursive.

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

Best Solution: BFS with Right-to-Left Traversal

Why BFS Right-to-Left Wins

The BFS with right-to-left traversal is the best for LeetCode 513 because it naturally processes the tree level by level, ensuring the last node we see in the deepest level is the leftmost one. It runs in O(n) time (n nodes) and O(w) space (w is max width), making it efficient and easy to follow. It’s like scanning the tree floor by floor, keeping the leftmost gem in sight!

How It Works

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

• Step 1: Use a Queue:
• Start with the root in a queue.
• Step 2: Traverse Right-to-Left:
• Process each level from right to left (append left child first, then right).
• Track the first node in the last level processed.
• Step 3: Return:
• The last level’s first node (leftmost) is the answer.
• Why It Works:
• BFS ensures level order.
• Right-to-left makes the leftmost node the last seen per level.

It’s like a leftmost spotlight sweep!

Step-by-Step Example

Example: [1,2,3,4,null,5,6,null,null,7]

• Tree:
``` 1 / \ 2 3 / / \ 4 5 6 / 7 ```
• Init: Queue = [1].
• Level 1:
• Dequeue 1, enqueue 3, 2 (right, left).
• Queue = [3, 2].
• Level 2:
• Dequeue 3, enqueue 6, 5.
• Dequeue 2, enqueue 4.
• Queue = [6, 5, 4].
• Level 3:
• Dequeue 6.
• Dequeue 5, enqueue 7.
• Dequeue 4.
• Queue = [7].
• Level 4:
• Dequeue 7, first node = 7.
• Queue empty, stop.
• Result: 7.

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 findBottomLeftValue(self, root: TreeNode) -> int:
        # Step 1: Initialize queue with root
        queue = deque([root])

        # Step 2: BFS traversal
        while queue:
            level_size = len(queue)
            # Process one level
            for i in range(level_size):
                node = queue.popleft()
                # Enqueue children right-to-left
                if node.right:
                    queue.append(node.right)
                if node.left:
                    queue.append(node.left)
            # Last level’s first node is leftmost
            if not queue:  # If this was the last level
                return node.val

        return root.val  # Single node case

• Line 11: Queue starts with root.
• Line 14: Loop until queue is empty.
• Line 15: Size of current level.
• Lines 17-21: Process level:
• Dequeue node.
• Add right, then left (ensures left is last processed).
• Line 23: If queue empty, node is last in bottom level.
• Line 26: Fallback for single node.
• Time Complexity: O(n)—visit each node once.
• Space Complexity: O(w)—queue holds max width.

It’s like a bottom-left radar!

Alternative Solution: DFS with Depth Tracking

Why an Alternative Approach?

The DFS with depth tracking uses recursion to explore the tree, tracking the deepest level and leftmost value at that depth. It’s O(n) time and O(h) space (h = height)—clever and compact but requires extra logic to prioritize left nodes. Great for recursion fans!

How It Works

Picture this as a depth-first dive:

• Step 1: Recursively traverse, tracking depth and max depth.
• Step 2: Update result when deeper and leftmost.
• Step 3: Return the leftmost value at max depth.

It’s like diving deep with a leftward bias!

Step-by-Step Example

Example: [2,1,3]

• Tree:
``` 2 / \ 1 3 ```
• Init: max_depth = -1, result = None.
• DFS:
• Left (1): Depth 1, max_depth = 1, result = 1.
• Right (3): Depth 1, same depth, no update (left first).
• Root (2): Depth 0, shallower, no update.
• Result: 1.

Code for DFS Approach

Copy codeclass Solution:
    def findBottomLeftValue(self, root: TreeNode) -> int:
        # Step 1: Variables to track max depth and result
        self.max_depth = -1
        self.result = None

        # Step 2: DFS helper
        def dfs(node, depth):
            if not node:
                return

            # Leaf node check
            if not node.left and not node.right:
                if depth > self.max_depth:
                    self.max_depth = depth
                    self.result = node.val
                return

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

        # Step 3: Start DFS
        dfs(root, 0)
        return self.result

• Lines 4-5: Track max depth and result.
• Lines 8-18: DFS:
• Base case: null node.
• Leaf: Update if deeper.
• Recurse left, then right (left priority).
• Line 21: Run DFS from root.
• Time Complexity: O(n)—visit each node.
• Space Complexity: O(h)—recursion stack.

It’s a depth-first left seeker!

Comparing the Two Solutions

• BFS Right-to-Left (Best):
• Pros: O(n) time, O(w) space, intuitive level order.
• Cons: Queue overhead.
• DFS with Depth (Alternative):
• Pros: O(n) time, O(h) space, compact.
• Cons: Recursive logic.

BFS wins for clarity!

Additional Examples and Edge Cases

• [1]: 1.
• ** [1,null,2]: 2.
• ** [1,2,null,3]: 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-headed champ!

Key Takeaways

• BFS: Level-by-level—learn at Python Basics!
• DFS: Depth with care.
• Trees: Bottom left is fun.
• Python: Queues or recursion, your pick!

Final Thoughts: Find That Bottom Left!

LeetCode 513: Find Bottom Left Tree Value in Python is a delightful tree challenge. BFS with right-to-left traversal is your clear path, while DFS adds depth-tracking flair. Want more? Try LeetCode 104: Maximum Depth of Binary Tree or LeetCode 515: Find Largest Value in Each Tree Row. Ready to explore? Head to Solve LeetCode 513 on LeetCode and uncover that bottom-left value today!