Solution Explanation for Binary Tree Vertical Order Traversal

This problem requires traversing a binary tree and returning a list of lists representing the vertical order traversal of the tree's nodes. The traversal should be from top to bottom, column by column, with nodes in the same row and column ordered from left to right.

Two approaches are presented: Depth-First Search (DFS) and Breadth-First Search (BFS). Both achieve the same result but differ in their traversal strategy and efficiency.

Approach 1: Depth-First Search (DFS)

This approach utilizes a recursive DFS function to explore the tree. For each node, we record its value, its depth, and its horizontal offset (column). The offset is calculated relative to a central column, with the root node at offset 0, left children having offsets decremented by 1, and right children having offsets incremented by 1.

The recorded data is stored in a dictionary (Python) or TreeMap (Java, C++) keyed by the horizontal offset. Values associated with each key are lists of (depth, value) tuples to maintain the vertical order within a column. After the DFS traversal, the dictionary is sorted by keys (offsets), and the inner lists are sorted by depth, resulting in the final vertical order traversal.

Time Complexity: O(N log N) The dominant factor is sorting the final result. The DFS traversal itself is O(N), but sorting the TreeMap or dictionary takes O(N log N) time in the worst case.

Space Complexity: O(N) to store the dictionary/TreeMap and the auxiliary recursion stack.

Approach 2: Breadth-First Search (BFS)

The BFS approach employs a queue to traverse the tree level by level. Similar to DFS, we maintain the horizontal offset for each node. However, nodes at the same level are processed together. A dictionary (Python) or TreeMap (Java, C++) stores the values of nodes with the same offset. Because BFS processes nodes level-by-level, the vertical ordering within a column is inherently maintained, removing the need for a secondary sort.

Time Complexity: O(N log N) The TreeMap operations (in Java and C++) have a logarithmic time complexity, resulting in a total time complexity of O(N log N).

Space Complexity: O(N) to store the queue and the dictionary/TreeMap.

Code Examples

The code examples for both DFS and BFS are provided in Python, Java, C++, and Go in the original response. The structure of the code is fairly consistent across languages. The choice of DFS or BFS influences mainly the traversal order and the need for post-traversal sorting. Note the use of defaultdict (Python), TreeMap (Java, C++), or equivalent data structures to efficiently store and manage the nodes by offset. The sorting process is also language-specific but follows the same principle.