#102

Medium

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Binary Tree Level Order Traversal

Perform a level order traversal on a binary tree using BFS to return node values level by level.

Problem Statement

Given the root of a binary tree, the task is to return the level order traversal of its nodes' values. This means you should traverse the tree from left to right, one level at a time, collecting all node values at each level.

For example, given a tree with root node value 3 and its children, the correct output should represent the nodes grouped by their respective levels. This problem requires efficient traversal using an appropriate approach like BFS, which can be achieved using a queue.

Examples

Example 1

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

Output: [[3],[9,20],[15,7]]

Example details omitted.

Example 2

Input: root = [1]

Output: [[1]]

Example details omitted.

Example 3

Input: root = []

Output: []

Example details omitted.

Constraints

The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000

Solution Approach

BFS with Queue

The most intuitive approach for this problem is using breadth-first search (BFS) with a queue. Start by enqueueing the root node. Then, for each level, dequeue nodes one by one, record their values, and enqueue their children (if any). This process ensures that the traversal happens level by level, left to right.

Queue Operations for Level Tracking

Using the queue to track nodes is crucial for managing the traversal levels. At each iteration, check the queue's size to determine how many nodes are at the current level. Process these nodes, then push their children into the queue for subsequent levels. This guarantees the BFS flow and keeps nodes grouped by their respective levels.

Edge Cases and Constraints

Handle edge cases where the tree is empty (i.e., no nodes). In such cases, return an empty list. Also, consider trees with only one node or a deep unbalanced tree. Since the number of nodes can be up to 2000, ensure that the algorithm performs efficiently within the problem's constraints.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is O(n), where n is the number of nodes in the binary tree, since we visit each node once. The space complexity is O(n) due to the queue used to store nodes level by level, which can grow up to the number of nodes at the widest level in the tree.

What Interviewers Usually Probe

Do you understand how to perform BFS traversal using a queue?
Can you explain why BFS is appropriate for this problem?
Are you able to handle edge cases like an empty tree or a tree with a single node?

Common Pitfalls or Variants

Common pitfalls

Misunderstanding the level order traversal concept and treating it as a depth-first traversal.
Failing to properly manage the queue to ensure that nodes are processed level by level.
Not accounting for edge cases such as an empty tree or single-node tree, which require special handling.

Follow-up variants

Spiral Level Order Traversal
Zigzag Level Order Traversal
Reverse Level Order Traversal

FAQ

What is the primary approach for solving Binary Tree Level Order Traversal?

The problem can be solved using breadth-first search (BFS) with a queue. This allows you to traverse nodes level by level.

How do we handle edge cases in Binary Tree Level Order Traversal?

Edge cases like an empty tree should return an empty list, while a tree with only one node should return that node in a list.

What is the time complexity of the Binary Tree Level Order Traversal solution?

The time complexity is O(n), where n is the number of nodes in the tree, as we process each node once during BFS.

Can you explain the space complexity for Binary Tree Level Order Traversal?

The space complexity is O(n), where n is the number of nodes in the tree, as we store nodes in the queue at each level.

What happens if we do not use BFS for Binary Tree Level Order Traversal?

Without BFS, we risk processing nodes in an incorrect order. BFS ensures nodes are processed level by level, maintaining the required order.

terminal

Solution

Solution 1: BFS

We can use the BFS method to solve this problem. First, enqueue the root node, then continuously perform the following operations until the queue is empty:

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# 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 levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            t = []
            for _ in range(len(q)):
                node = q.popleft()
                t.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            ans.append(t)
        return ans

Continue Practicing

#101 symmetric tree #103 binary tree zigzag level order traversal #100 same tree