Skip to content

102. Binary Tree Level Order Traversal

Medium

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

Solution

from collections import deque


class Solution:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if root is None:
            return None

        res = []
        queue = deque([root])

        while len(queue) > 0:
            level = []
            for _ in range(len(queue)):
                curr = queue.popleft()

                level.append(curr.val)
                if curr.left is not None:
                    queue.append(curr.left)
                if curr.right is not None:
                    queue.append(curr.right)

            res.append(level)

        return res


class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right