LeetCode 314: Binary Tree Vertical Order Traversal Solution in Python – A Step-by-Step Guide

What Is LeetCode 314: Binary Tree Vertical Order Traversal?

In LeetCode 314: Binary Tree Vertical Order Traversal, you’re given a binary tree, and your task is to return a list of lists where each sublist contains the node values in a vertical column, ordered from top to bottom and left to right across columns. For example, in a tree with root 3, left 9, right 20 (with 20 having children 15 and 7), the vertical order is [[9], [3, 15], [20], [7]]. This problem tests your ability to navigate a tree while tracking horizontal positions, building on classics like LeetCode 102: Binary Tree Level Order Traversal.

Problem Statement

• Input: A binary tree root node.
• Output: A list of lists—node values by vertical column, top-to-bottom within each, left-to-right across columns.
• Rules:
• Columns are indexed: root at 0, left child at -1, right child at +1, etc.
• Within a column, order is top-down.

Constraints

• Number of nodes: 0 to 100
• -100 <= Node.val <= 100

Examples

• Input: root = [3,9,20,null,null,15,7]
• Output: [[9],[3,15],[20],[7]]
• Input: root = [1,2,3,4,5,6,7]
• Output: [[4],[2],[1,5,6],[3],[7]]
• Input: root = []
• Output: []

Understanding the Problem: Seeing Vertically

To solve LeetCode 314: Binary Tree Vertical Order Traversal in Python, we need to traverse the tree, assign each node a column index, and group values by column while preserving top-down order within each. A naive approach might traverse without structure, but we need order. We’ll use:

• Best Solution (BFS): O(n) time, O(n) space—level-by-level ensures top-down order.
• Alternative Solution (DFS): O(n) time, O(n) space—depth-first with extra sorting.

Let’s start with the BFS solution, explained in a beginner-friendly way.

Best Solution: BFS with Column Tracking

Why This Is the Best Solution

The BFS approach is the top choice for LeetCode 314 because it’s efficient—O(n) time, O(n) space—and naturally preserves the top-down order within each column by processing level-by-level. It uses a queue to traverse and a dictionary to group nodes by column, making it intuitive and fast. It’s like scanning the tree from top to bottom, column by column—perfect for the task!

How It Works

Think of this as a sideways sweep:

• Step 1: Initialize:
• Use a queue with (node, column) pairs, starting at (root, 0).
• Use a dictionary to store column → list of values.
• Step 2: Traverse Level-by-Level:
• Pop node, add its value to its column’s list.
• Push left child (col-1) and right child (col+1).
• Step 3: Collect Results:
• Sort columns, return their value lists.
• Step 4: Why It Works:
• BFS ensures top-down order within columns.
• Dictionary handles column grouping efficiently.

It’s like a tree X-ray—clear and orderly!

Step-by-Step Example

Example: root = [3,9,20,null,null,15,7]

• Init: Queue = [(3,0)], Cols = {}
• Level 1: Pop (3,0), Cols = {0:[3]}, Push (9,-1), (20,1)
• Level 2:
• Pop (9,-1), Cols = {-1:[9], 0:[3]}
• Pop (20,1), Cols = {-1:[9], 0:[3], 1:[20]}, Push (15,0), (7,2)
• Level 3:
• Pop (15,0), Cols = {-1:[9], 0:[3,15], 1:[20]}
• Pop (7,2), Cols = {-1:[9], 0:[3,15], 1:[20], 2:[7]}
• Result: Sort keys: [[9], [3,15], [20], [7]]

Code with Detailed Line-by-Line Explanation

from collections import defaultdict, deque

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

        # Queue for BFS: (node, column)
        queue = deque([(root, 0)])
        # Dictionary: column -> list of values
        cols = defaultdict(list)

        # BFS traversal
        while queue:
            node, col = queue.popleft()
            cols[col].append(node.val)
            if node.left:
                queue.append((node.left, col - 1))
            if node.right:
                queue.append((node.right, col + 1))

        # Sort columns and return values
        return [cols[col] for col in sorted(cols.keys())]

• Lines 5-6: Handle empty tree.
• Line 8: Start queue with root at column 0.
• Line 10: Use defaultdict for column lists.
• Lines 13-18: BFS:
• Pop node, add value to its column.
• Push children with adjusted columns.
• Line 20: Sort column keys, return lists.
• Time Complexity: O(n)—visit each node once.
• Space Complexity: O(n)—queue and dictionary.

This is like a tree-column scanner—smooth and fast!

Alternative Solution: DFS with Column and Row Tracking

Why an Alternative Approach?

The DFS approach also runs in O(n) time, O(n) space, but uses depth-first traversal, requiring extra tracking of row positions to ensure top-down order. It’s a good alternative for understanding tree traversal trade-offs, though it needs sorting within columns. It’s like exploring the tree’s depths first—insightful but more complex!

How It Works

Dive deep with DFS:

• Step 1: Track (col, row, val) for each node.
• Step 2: Use DFS to visit all nodes, storing in a list.
• Step 3: Group by column, sort by row within each.

Step-by-Step Example

Example: root = [1,2,3]

• DFS:
• (0,0,1), (1,1,3), (-1,1,2)
• List: [(0,0,1), (-1,1,2), (1,1,3)]
• Group: {-1:[(1,2)], 0:[(0,1)], 1:[(1,3)]}
• Result: [[2], [1], [3]]

Code for DFS Approach

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

        # List to store (col, row, val)
        nodes = []

        def dfs(node, row, col):
            if not node:
                return
            nodes.append((col, row, node.val))
            dfs(node.left, row + 1, col - 1)
            dfs(node.right, row + 1, col + 1)

        # Traverse tree
        dfs(root, 0, 0)

        # Group by column, sort by row
        cols = defaultdict(list)
        for col, row, val in sorted(nodes):
            cols[col].append(val)

        return [cols[col] for col in sorted(cols.keys())]

• Time Complexity: O(n)—visit each node, plus O(n log n) sorting.
• Space Complexity: O(n)—nodes list and dictionary.

It’s a deep tree explorer!

Comparing the Two Solutions

• BFS (Best):
• Pros: O(n) time, O(n) space, natural top-down order.
• Cons: Queue management.
• DFS (Alternative):
• Pros: O(n) time, O(n) space, recursive simplicity.
• Cons: Extra sorting needed.

BFS wins for simplicity and order.

Additional Examples and Edge Cases

• [1]: [[1]].
• [1,2]: [[2], [1]].
• []: [].

Both handle these perfectly.

Complexity Breakdown

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

BFS is faster overall.

Key Takeaways

• BFS: Level-by-level—perfect fit!
• DFS: Depth-first—educational twist!
• Python: Queues and dicts shine—see [Python Basics](/python/basics).
• Trees: Vertical views are cool.

Final Thoughts: See the Tree Sideways

LeetCode 314: Binary Tree Vertical Order Traversal in Python is a tree traversal gem. BFS offers efficiency and natural order, while DFS provides a recursive angle. Want more tree adventures? Try LeetCode 102: Binary Tree Level Order Traversal or LeetCode 199: Binary Tree Right Side View. Ready to traverse? Head to Solve LeetCode 314 on LeetCode and scan those columns today!