Binary Tree Level Order Traversal

Problem

https://leetcode.com/problems/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

Pattern

Tree, Breadth-First Search, Binary Tree

Approaches

BFS

Explanation

A level order traversal is a breadth-first traversal of a binary tree. Track all the nodes in the current level and add their values to the traversal. At the same time, add their children to the next level. We finish traversing once all the chilren are None.

Code

class TreeNode:
    """Node in a binary tree."""

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


def levelOrder(root: TreeNode | None) -> list[list[int]]:
    """Return the level order traversal of a binary tree."""
    traversal = []
    level = [root] if root is not None else []
    while len(level) > 0:
        values = []
        next_level = []
        for node in level:
            values.append(node.val)
            if node.left is not None:
                next_level.append(node.left)
            if node.right is not None:
                next_level.append(node.right)

        traversal.append(values)
        level = next_level

    return traversal

Test

>>> from binary_tree_level_order_traversal__bfs import levelOrder, TreeNode
>>> root = TreeNode(3, TreeNode(9), TreeNode(20, TreeNode(15), TreeNode(7)))
>>> levelOrder(root)
[[3], [9, 20], [15, 7]]
>>> root = TreeNode(1)
>>> levelOrder(root)
[[1]]
>>> root = None
>>> levelOrder(root)
[]

Complexity

\(n\) is the number of nodes in the binary tree

Measure	Complexity	Notes
Time	\(O(n)\)	visit each node once
Auxiliary Space	\(O(n)\)	next_level holds one level of nodes, up to \(O(n / 2)\) for a full binary tree

class binary_tree_level_order_traversal__bfs.TreeNode(val, left=None, right=None)

Bases: object

Node in a binary tree.

binary_tree_level_order_traversal__bfs.levelOrder(root: TreeNode | None) → list[list[int]]

Return the level order traversal of a binary tree.