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). (Problem)

				

    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 []
            
            q = deque([root])
            sol=[[root.val]]
            
            while q:
                a=[]
                for i in range(len(q)):
                    
                    node = q.popleft()
                    
                    if node.left:
                        q.append(node.left)
                        a.append(node.left.val)
                    if node.right:
                        q.append(node.right)
                        a.append(node.right.val)
                if len(a)!=0:        
                    sol.append(a)
            return sol
							
			

Time Complexity => o(n)

Space Complexity => o(n) (height of tree)