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).

Idea

BFS

# 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 []
        ans = []
        d = deque([root])
        while d:
            level_nodes = []
            for _ in range(len(d)):
                element = d.popleft()
                level_nodes.append(element.val)
                if element.left:
                    d.append(element.left)
                if element.right:
                    d.append(element.right)
            ans.append(level_nodes)
        return ans