Leetcode #987: Vertical Order Traversal of a Binary Tree
In this guide, we solve Leetcode #987 Vertical Order Traversal of a Binary Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Hash Table, Binary Tree, Sorting
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def verticalTraversal(self, root: Optional[TreeNode]) -> List[List[int]]:
def dfs(root: Optional[TreeNode], i: int, j: int):
if root is None:
return
nodes.append((j, i, root.val))
dfs(root.left, i + 1, j - 1)
dfs(root.right, i + 1, j + 1)
nodes = []
dfs(root, 0, 0)
nodes.sort()
ans = []
prev = -2000
for j, _, val in nodes:
if prev != j:
ans.append([])
prev = j
ans[-1].append(val)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.