How do you solve Binary Tree Level Order Traversal on LeetCode?

Binary Tree Level Order Traversal (LeetCode #102) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a queue allows for efficient level-wise traversal.

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

The optimal solution for Binary Tree Level Order Traversal runs in O(n) time and uses O(n) space. The time complexity is O(n) because we visit each node exactly once. The space complexity is O(n) due to the queue storing nodes at each level.

What is the brute force solution for Binary Tree Level Order Traversal?

The brute force approach for Binary Tree Level Order Traversal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves traversing the tree level by level and collecting nodes' values into lists for each level. This method is straightforward but inefficient, as it may require multiple passes through the tree. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Binary Tree Level Order Traversal?

Binary Tree Level Order Traversal uses the following concepts: Tree, Breadth-First Search, Binary Tree. The recommended patterns to study are: Breadth-First Search, Queue.

What are the interview tips for Binary Tree Level Order Traversal?

Always clarify if the tree can be empty. Explain your thought process while coding. Consider edge cases like single-node trees or completely unbalanced trees.

What are common mistakes when solving Binary Tree Level Order Traversal?

Not handling null nodes properly in the queue. Confusing depth-first search with breadth-first search.

#102

Binary Tree Level Order Traversal

Medium

TreeBreadth-First SearchBinary TreeBreadth-First SearchQueue

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach uses a queue to perform a breadth-first search (BFS) on the tree. This allows us to process each level of the tree in a single pass, making it much more efficient.

⚙️

Algorithm

4 steps

1Step 1: Initialize a queue and add the root node to it.
2Step 2: While the queue is not empty, determine the number of nodes at the current level.
3Step 3: Dequeue each node, add its value to the current level list, and enqueue its children.
4Step 4: After processing all nodes at the current level, add the level list to the result.

solution.py25 lines

1# Full working Python code
2from collections import deque
3class TreeNode:
4    def __init__(self, val=0, left=None, right=None):
5        self.val = val
6        self.left = left
7        self.right = right
8
9def levelOrder(root):
10    if not root:
11        return []
12    result = []
13    queue = deque([root])
14    while queue:
15        level_size = len(queue)
16        level = []
17        for _ in range(level_size):
18            node = queue.popleft()
19            level.append(node.val)
20            if node.left:
21                queue.append(node.left)
22            if node.right:
23                queue.append(node.right)
24        result.append(level)
25    return result

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(n) due to the queue storing nodes at each level.

1Using a queue allows for efficient level-wise traversal.
2Understanding tree depth helps in visualizing the traversal.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.