LeetCode - Binary Tree Level Order Traversal

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: []

 

Constraints:

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

 

My Solution:

# 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]]:
        if not root:
            return []
        
        level_list = []

        def levelTraverse(root, level):
            try:
                level_list[level].append(root.val)
            except IndexError:
                level_list.append([root.val])
            
            if root.left:
                levelTraverse(root.left, level + 1)
            if root.right:
                levelTraverse(root.right, level + 1)
        
        levelTraverse(root, 0)

        return level_list

 

 

ChatGPT's Solution:

from collections import deque


# 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]]:
        if not root:
            return []
        
        res = []
        queue = deque([root])

        while queue:
            level = []
            for _ in range(len(queue)):  # Process all nodes at the current level
                node = queue.popleft()
                level.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            res.append(level)
        
        return res