102. Binary Tree Level Order Traversal

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: []

Constraints:

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

💡 Hint 1:
Use a queue to perform BFS.

Submitted Code

# 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]]:
        result = []
        if not root:
            return result
        
        q = deque()
        q.append(root)

        while q:
            level = []
            for _ in range(len(q)):
                node = q.popleft()
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
                level.append(node.val)
            result.append(level)

        return result

Runtime: 0 ms | Beats 100.00%
Memory: 19.93 MB | Beats 60.85%

Other Solutions

1st

class Solution:
    def levelOrder(self, root):
        if not root:
            return []
        ans, level = [], [root]
        while level:
            ans.append([node.val for node in level])
            temp = []
            for node in level:
                temp.extend([node.left, node.right])
            level = [leaf for leaf in temp if leaf]
        return ans

time complexity: 𝑂(𝑛)
space complexity: 𝑂(𝑛)

큐 없이 리스트만으로 BFS를 구현한 방법이다.