LeetCode 101: 102. Binary Tree Level Order Traversal

LeetCode Problem: 102. Binary Tree Level Order Traversal

https://leetcode.com/problems/binary-tree-level-order-traversal/description/

Problem Description:

Given the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).

Example 1:

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

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Original Code (BFS Approach):

from collections import deque

class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []  # If the tree is empty, return an empty list

        res = []  # Initialize the result list
        queue = deque([root])  # Initialize the queue with the root

        while queue:  # While there are nodes to process
            level_size = len(queue)  # Number of nodes at the current level
            level_nodes = []  # List to hold nodes at the current level

            for _ in range(level_size):  # Process all nodes at the current level
                node = queue.popleft()  # Dequeue the front node
                level_nodes.append(node.val)  # Add the node's value to the current level's list

                # Enqueue the left and right children, if they exist
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)

            res.append(level_nodes)  # Add the current level's values to the result

        return res  # Return the final result

Line-by-Line Explanation and Example Execution:

from collections import deque
- Imports deque for efficient queue operations.
class Solution:
- Defines the Solution class.
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
- Defines the main function levelOrder, which takes the root of a binary tree and returns a list of lists containing the level order traversal.
if not root:
- Checks if the tree is empty. If so, returns an empty list.
res = []
- Initializes an empty list to store the result of the level order traversal.
queue = deque([root])
- Initializes a queue with the root node to keep track of nodes to be processed.
while queue:
- Continues processing as long as there are nodes in the queue.
level_size = len(queue)
- Stores the number of nodes at the current level.
level_nodes = []
- Initializes a list to hold the values of nodes at the current level.
for _ in range(level_size):
- Iterates through all nodes at the current level.
node = queue.popleft()
- Dequeues the front node from the queue.
level_nodes.append(node.val)
- Adds the value of the current node to the list of nodes at the current level.
if node.left:
- Checks if the left child exists. If so, enqueues it.
if node.right:
- Checks if the right child exists. If so, enqueues it.
res.append(level_nodes)
- Adds the current level’s values to the result list.
return res
- Returns the final result after processing all levels.

Example Execution:

Assuming we have the following binary tree:

Execution steps:

Initialization:
- root = 3
- res = [], queue = deque([3])
First Level:
- level_size = 1, level_nodes = []
- Dequeue 3, add to level_nodes → level_nodes = [3]
- Enqueue 9 and 20.
- Add level_nodes to res → res = [[3]].
Second Level:
- level_size = 2, level_nodes = []
- Dequeue 9, add to level_nodes → level_nodes = [9]
- Dequeue 20, add to level_nodes → level_nodes = [9, 20]
- Enqueue 15 and 7.
- Add level_nodes to res → res = [[3], [9, 20]].
Third Level:
- level_size = 2, level_nodes = []
- Dequeue 15, add to level_nodes → level_nodes = [15]
- Dequeue 7, add to level_nodes → level_nodes = [15, 7]
- Add level_nodes to res → res = [[3], [9, 20], [15, 7]].
Final Result:
- Return res = [[3], [9, 20], [15, 7]].

Time Complexity Analysis:

Time Complexity: O(n)
- Where n is the number of nodes in the tree. Each node is processed once.

Space Complexity Analysis:

Space Complexity: O(n)
- In the worst case, the queue can hold up to n nodes (in case of a complete binary tree).

Tips and Warnings:

Edge Cases:
- Consider cases where the tree is empty or consists of a single node.
Understanding BFS:
- Ensure understanding of breadth-first search and how it explores levels in the tree.
Level Order Traversal:
- Familiarity with how to implement level order traversal will help in solving similar problems.

Summary

BFS Method: Effectively performs level order traversal of a binary tree, with a time complexity of O(n) and a space complexity of O(n).
Clear and Understandable: The code is straightforward and easy to understand, suitable for handling this type of problem.

Application Tips

Choose Appropriate Methods:
- Select the method best suited to specific problem requirements to ensure efficiency and readability.
Handle Edge Cases:
- Always consider edge cases in algorithm design to avoid potential errors.
Optimize Space Usage:
- When handling large datasets, consider using space-efficient algorithms.

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