Binary Tree Vertical Order Traversal

Question (LC.314)

Given a binary tree, return the vertical order traversal of its nodes' values.

Example

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

I: root = [3,9,8,4,0,1,7,null,null,null,2,5]
O: [[4], [9,5], [3,0,1], [8,2], [7]]

Analysis

The key of unlocking this question is to understand the definition of vertical order.

i.e. from top to bottom, column by column

if two nodes are in the same row and column, the order should be from left to right.

There are two conditions in this ordering, rows and columns. I have missed the rows part when I first read this question.

For example, [3,9,8,4,0,1,7,null,null,null,2,5] has the following diagram.

Vertical order binary tree

Which can be represented in a matrix like this

row/col	-2	-1	0	1	2
0			3
1		9		8
2	4		0,1		7
3		5		2

A corner case without row key with DFS is [8, 2] will be [2, 8]. BFS gets row key sort of for free.

DFS Code

class Solution:
    
    def verticalOrder(self, root: TreeNode) -> List[List[int]]:
        
        results: List[List[int]] = []
        
        if root is None:
            return results 
        
        node_dict: Dict[TreeNode, Tuple[int, int]] = {}
        
        self.preorder(root, node_dict, 0, 0)
        
        # group by col score 
        score_dict: Dict[int, List[Tuple[int, int]]] = defaultdict(list)
        
        for node in node_dict:
            col_score = node_dict[node][0]
            row_score = node_dict[node][1]
            score_dict[col_score].append((node.val, row_score))
        
        sorted_scores = sorted(score_dict)
        
        for score in sorted_scores:
            # sort by row score 
            vertical_order = [
                ele[0] for ele in sorted(score_dict[score], key=lambda t: t[1])
            ]
            results.append(
                vertical_order
            )
        
        return results
     
        
    def preorder(
        self, 
        node: TreeNode, 
        node_dict: Dict[TreeNode, Tuple[int, int]], 
        col_score: int,
        row_score: int
    ) -> None:
        
        if node is None:
            return 
        
        # visit 
        node_dict[node] = (col_score, row_score)
        
        # go left 
        self.preorder(node.left, node_dict, col_score - 1, row_score + 1) 
        
        # go right 
        self.preorder(node.right, node_dict, col_score + 1, row_score + 1)
    

time O(nlogn)

space O(n)

BFS Code

def verticalOrder(self, root: TreeNode) -> List[List[int]]:
    
    results: List[List[int]] = []
    
    if root is None:
        return results 
    
    queue: Deque[Tuple[TreeNode, int]] = deque()
    queue.append((root, 0))
    col_map: Dict[int, List[int]] = defaultdict(list)
    
    while len(queue) > 0:
        cur_tuple = queue.popleft()
        cur_node = cur_tuple[0]
        cur_col = cur_tuple[1]
        col_map[cur_col].append(cur_node.val)
        
        if cur_node.left:
            queue.append((cur_node.left, cur_col - 1))
        
        if cur_node.right:
            queue.append((cur_node.right, cur_col + 1))
    
    sorted_keys = sorted(col_map)
    
    for key in sorted_keys:
        results.append(col_map[key])
    
    return results 

time O(nlogn)

space O(n)