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Vertical Order Traversal of Binary Tree in DSA Rust - Time & Space Complexity

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Time Complexity: Vertical Order Traversal of Binary Tree
O(n log k)
Understanding Time Complexity

We want to understand how the time needed grows when we do a vertical order traversal of a binary tree.

Specifically, how the number of nodes affects the work done to group nodes by their vertical columns.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.


use std::collections::{BTreeMap, VecDeque};

fn vertical_order(root: Option<&Box>) -> Vec> {
    let mut map: BTreeMap> = BTreeMap::new();
    let mut queue: VecDeque<(&Box, i32)> = VecDeque::new();
    if let Some(node) = root {
        queue.push_back((node, 0));
    }
    while let Some((node, col)) = queue.pop_front() {
        map.entry(col).or_default().push(node.val);
        if let Some(left) = &node.left {
            queue.push_back((left, col - 1));
        }
        if let Some(right) = &node.right {
            queue.push_back((right, col + 1));
        }
    }
    map.values().cloned().collect()
}

This code visits each node in the tree once, grouping node values by their vertical column index.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: The while loop that visits each node once.
  • How many times: Exactly once per node, so n times for n nodes.
How Execution Grows With Input

As the number of nodes grows, the code visits each node once and adds its value to a map.

Input Size (n)Approx. Operations
10About 10 visits and insertions
100About 100 visits and insertions
1000About 1000 visits and insertions

Pattern observation: The work grows linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n log k)

This means the time grows a bit more than linearly because inserting into the map takes extra time, but overall it scales well with the number of nodes.

Common Mistake

[X] Wrong: "The traversal is just O(n) because we visit each node once, so no extra cost."

[OK] Correct: While visiting nodes is O(n), inserting into a sorted map (BTreeMap) takes O(log k) per insertion, making total time O(n log k), where k is the number of distinct columns.

Interview Connect

Understanding this complexity helps you explain how tree traversals combined with data structures affect performance, a common topic in interviews.

Self-Check

What if we used a hash map instead of a BTreeMap for grouping columns? How would the time complexity change?

Practice

(1/5)
1. What does vertical order traversal of a binary tree mean?
easy
A. Nodes are grouped by their vertical columns from left to right
B. Nodes are visited level by level from top to bottom
C. Nodes are visited in preorder (root, left, right)
D. Nodes are visited in inorder (left, root, right)

Solution

  1. Step 1: Understand vertical order traversal

    Vertical order traversal groups nodes by their vertical position (column) relative to the root.

  2. Step 2: Compare with other traversals

    Unlike level order or inorder, vertical order focuses on columns from left to right.

  3. Final Answer:

    Nodes are grouped by their vertical columns from left to right -> Option A

  4. Quick Check:

    Vertical order = group by vertical columns [OK]
Hint: Think columns, not levels or order [OK]
Common Mistakes:
  • Confusing vertical order with level order traversal
  • Thinking vertical order is preorder or inorder
  • Ignoring horizontal distance from root
2. Which Rust data structure is best to store nodes grouped by their vertical column during vertical order traversal?
easy
A. Vec<i32>
B. HashMap
C. BinaryHeap<i32>
D. LinkedList<i32>

Solution

  1. Step 1: Identify grouping need

    We need to group nodes by their vertical column index (an integer key).

  2. Step 2: Choose suitable Rust structure

    HashMap maps column indices to lists of node values, perfect for grouping.

  3. Final Answer:

    HashMap -> Option B

  4. Quick Check:

    Use HashMap for key-value grouping [OK]
Hint: Use HashMap for key-based grouping [OK]
Common Mistakes:
  • Using Vec alone without keys
  • Using BinaryHeap which is for priority queues
  • Using LinkedList which is inefficient here
3. Given the binary tree: 

    1
   / \
  2   3
 /     \
4       5
What is the vertical order traversal output?
medium
A. [[4], [2], [1], [3], [5]]
B. [[1], [2, 3], [4, 5]]
C. [[4, 2], [1, 3], [5]]
D. [[5], [3], [1], [2], [4]]

Solution

  1. Step 1: Assign vertical columns

    Root 1 is column 0; left child 2 is -1; left child of 2 (4) is -2; right child 3 is +1; right child of 3 (5) is +2.

  2. Step 2: Group nodes by columns

    Column -2: [4], -1: [2], 0: [1], 1: [3], 2: [5].

  3. Final Answer:

    [[4], [2], [1], [3], [5]] -> Option A

  4. Quick Check:

    Group nodes by horizontal distance [OK]
Hint: Track horizontal distance from root [OK]
Common Mistakes:
  • Mixing vertical order with level order
  • Grouping nodes incorrectly by level
  • Ignoring negative column indices
4. Consider this Rust code snippet for vertical order traversal: 

fn vertical_order(root: Option) -> Vec {
    let mut map = std::collections::HashMap::new();
    fn dfs(node: &Option, col: i32, map: &mut HashMap) {
        if let Some(n) = node {
            map.entry(col).or_insert(vec![]).push(n.val);
            dfs(&n.left, col - 1, map);
            dfs(&n.right, col + 1, map);
        }
    }
    dfs(&root, 0, &mut map);
    map.values().cloned().collect()
}
What is the main bug causing incorrect output?
medium
A. DFS should use BFS for vertical order
B. map.entry() should use get_mut() instead
C. col variable should start at 1, not 0
D. map.values() does not guarantee sorted column order

Solution

  1. Step 1: Analyze map usage

    HashMap stores columns but values() returns in arbitrary order, not sorted by column.

  2. Step 2: Understand output order requirement

    Vertical order requires columns sorted from left to right, so keys must be sorted before collecting values.

  3. Final Answer:

    map.values() does not guarantee sorted column order -> Option D

  4. Quick Check:

    Sort keys before output [OK]
Hint: Sort keys before collecting values [OK]
Common Mistakes:
  • Assuming HashMap preserves order
  • Thinking BFS is mandatory for vertical order
  • Starting column index incorrectly
5. You want to modify vertical order traversal to also sort nodes by their depth (top to bottom) within the same vertical column. Which approach is best?
hard
A. Traverse tree in preorder and collect nodes without sorting
B. Use a Vec indexed by column and append nodes as visited
C. Use a HashMap> and sort each vector by depth before output
D. Use a BinaryHeap keyed by node value only

Solution

  1. Step 1: Understand sorting by depth need

    Nodes in same vertical column must be ordered by their depth (distance from root).

  2. Step 2: Choose data structure to store depth info

    Store (depth, value) pairs in HashMap for each column, then sort by depth before output.

  3. Final Answer:

    Use a HashMap> and sort each vector by depth before output -> Option C

  4. Quick Check:

    Store depth info and sort before output [OK]
Hint: Keep depth info to sort nodes in columns [OK]
Common Mistakes:
  • Ignoring depth and outputting unsorted nodes
  • Using BinaryHeap without depth key
  • Appending nodes without sorting
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