Binary Tree Level Order

Interview Problem Intermediate Medium Binary Tree Queue Breadth-first Search

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Problem

Implement Solution.levelOrder(root). Return one list of values for each depth from left to right. Return an empty list for an empty tree.

Starter code

class Solution:
    def levelOrder(self, root):
        pass

Test cases

three-levels

{
  "args": [
    {
      "$type": "binary-tree",
      "values": [
        3,
        9,
        20,
        null,
        null,
        15,
        7
      ]
    }
  ]
}

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

Wizard outline

Step 1: Initialize Solution.levelOrder
Replace the empty starter with the first real state owned by Solution.levelOrder. A small, named state is easier to verify than a complete algorithm. Establish it before adding the branch or loop that changes it.
Step 2: Pass the Empty case
Complete the readable core algorithm for one representative Interview case. Use the queue size at the start of each round to define one tree level.
Step 3: Harden the Three Levels boundary
Repair the reviewed boundary and pass the complete submission contract. The queue initially contains exactly the current level; consuming that fixed count records every node at that depth while enqueuing precisely the next level.

Footguns and prerequisites

Do not let children appended during a round extend that same level.
trees and graphs

Reviewed references

LeetCode 102 — Binary Tree Level Order Traversal

Practice prerequisites

Expand One Tree Level(opens in a new tab)
Expand One Tree Level isolates at the start of each outer iteration, the queue contains exactly the current level in left-to-right order. That focused state discipline is required when implementing binary tree level order as a complete Interview Problem.

Recommended approach and implementation

Run breadth-first search with a deque and consume exactly the queue length captured at the start of each level.

Why it works: The queue initially contains exactly the current level; consuming that fixed count records every node at that depth while enqueuing precisely the next level.

from collections import deque

class Solution:
    def levelOrder(self, root):
        """
        Checkpoint 1: initialize the state owned by this Interview contract.
        Checkpoint 2: assemble the primary transition without hiding the boundary.
        """
        if root is None:
            return []
        result = []
        queue = deque([root])
        while queue:
            level = []
            for _ in range(len(queue)):
                node = queue.popleft()
                level.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            result.append(level)
        return result