Introduction

In this blog post, we will explore how to solve the LeetCode problem "987" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given the root of a binary tree, calculate the vertical order traversal of the binary tree. For each node at position (row, col), its left and right children will be at positions (row + 1, col - 1) and (row + 1, col + 1) respectively. The root of the tree is at (0, 0). The vertical order traversal of a binary tree is a list of top-to-bottom orderings for each column index starting from the leftmost column and ending on the rightmost column. There may be multiple nodes in the same row and same column. In such a case, sort these nodes by their values. Return the vertical order traversal of the binary tree. Example 1: Input: root = [3,9,20,null,null,15,7] Output: [[9],[3,15],[20],[7]] Explanation: Column -1: Only node 9 is in this column. Column 0: Nodes 3 and 15 are in this column in that order from top to bottom. Column 1: Only node 20 is in this column. Column 2: Only node 7 is in this column. Example 2: Input: root = [1,2,3,4,5,6,7] Output: [[4],[2],[1,5,6],[3],[7]] Explanation: Column -2: Only node 4 is in this column. Column -1: Only node 2 is in this column. Column 0: Nodes 1, 5, and 6 are in this column. 1 is at the top, so it comes first. 5 and 6 are at the same position (2, 0), so we order them by their value, 5 before 6. Column 1: Only node 3 is in this column. Column 2: Only node 7 is in this column. Example 3: Input: root = [1,2,3,4,6,5,7] Output: [[4],[2],[1,5,6],[3],[7]] Explanation: This case is the exact same as example 2, but with nodes 5 and 6 swapped. Note that the solution remains the same since 5 and 6 are in the same location and should be ordered by their values. Constraints: The number of nodes in the tree is in the range [1, 1000]. 0 <= Node.val <= 1000

Explanation

• Breadth-First Search (BFS): Traverse the tree level by level using BFS to maintain the top-to-bottom order.
  • Column Mapping: Use a hash map (dictionary in Python) to store nodes based on their column index. Update the column index as we move left (col - 1) or right (col + 1).
  • Sorting: Sort nodes within each column based on their row and value, handling cases where multiple nodes occupy the same position.

• Time Complexity: O(W H log(H)), where W is the width of the tree and H is the height. The W * H comes from visiting all nodes, and the log(H) comes from sorting the nodes in each column based on their row and value.
• Space Complexity: O(N), where N is the number of nodes in the tree, to store the column mapping and the queue for BFS.

Code

    from collections import defaultdict, deque
from typing import List, Optional


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


def vertical_order(root: Optional[TreeNode]) -> List[List[int]]:
    if not root:
        return []

    column_map = defaultdict(list)
    queue = deque([(root, 0, 0)])  # (node, column, row)

    while queue:
        node, column, row = queue.popleft()
        column_map[column].append((row, node.val))

        if node.left:
            queue.append((node.left, column - 1, row + 1))
        if node.right:
            queue.append((node.right, column + 1, row + 1))

    sorted_columns = sorted(column_map.keys())
    result = []
    for col in sorted_columns:
        column_map[col].sort()  # Sort by row, then value
        result.append([val for row, val in column_map[col]])

    return result