A-053● Day 532026.06.29mediumleetcode #102neetcode150

Binary Tree Level Order Traversal

#tree#breadth-first-search#binary-tree

↗ leetcode #102 · binary-tree-level-order-traversal

Problem

· problem

P.053

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).

constraints

· 0 ≤ number of nodes ≤ 2000
· -1000 ≤ Node.val ≤ 1000

// paraphrased summary — see source for full text

examples

example 1input → output

[3,9,20,null,null,15,7]

[[3],[9,20],[15,7]]

example 2input → output

[1]

[[1]]

example 3input → output

[]

[]

Pre-solve

· pre-solve

● 1list shown→● 2select→● 3reveal

☐
What does "level order traversal" mean exactly?
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Should nodes within each level be ordered left-to-right?
☐
What should we return for an empty tree?
☐
Can node values be negative?
☐
Should we return nodes in a single flattened list?
☐
Would depth-first search be a better approach?

check the items you would ask, then press confirm

// session-only state — refresh resets (repeatable practice)

Logic Structure

· logic

● 1slots shown→● 2pick per slot→● 3reveal

// pick one code line per slot to assemble the algorithm flow. no typing — just the logic skeleton.

step 1· Initialize result list

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res = []

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res = [[]]

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res = None

step 2· Initialize queue

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q = collections.deque()

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q = []

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q = Stack()

step 3· Seed queue with root

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q.append(root)

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q.append(root.val)

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res.append([root.val])

step 4· Process while queue has nodes

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while q:

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while q is not None:

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while len(q) > 0:

step 5· Process exactly current level's nodes│ nested

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for i in range(len(q)):

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for node in q:

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for i in range(len(q) - 1):

step 6· Dequeue and record node value│ │ nested

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node = q.popleft()

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node = q.pop()

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node = q[0]

pick one option per slot

// format: slot — recursive / DP patterns use ordering / state-first formats. ADR-08 follow-up.

Solve · Trace

· solve

solution.py

# Definition for a binary tree node.

# class TreeNode:

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution:

    def levelOrder(self, root: TreeNode) -> List[List[int]]:

        res = []

        q = collections.deque()

        if root:

            q.append(root)

        while q:

            val = []

            for i in range(len(q)):

                node = q.popleft()

                val.append(node.val)

                if node.left:

                    q.append(node.left)

                if node.right:

                    q.append(node.right)

            res.append(val)

        return res

mental dry-run cases

// walk each case in your head; expand the worked example below if stuck.

case 1

[3,9,20,null,null,15,7]

→

[[3],[9,20],[15,7]]

case 2

[1]

→

[[1]]

case 3

[]

→

[]

// UI does not walk-through — you do the dry-run mentally. Expand the worked example if stuck.