Binary Tree Level Order Traversal

🧩 Problem Statement

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]]

🔍 Approaches

1. Breadth-First Search (BFS) ($O(N)$)

• Concept: Use a queue to traverse the tree level by level.
• Algorithm:

1. If root is null, return empty.
2. Initialize a queue with root.
3. While queue is not empty:

• Get the number of nodes at the current level (size = q.size()).
• Iterate size times:
• Pop a node.
• Add its value to the current level list.
• Push its valid children (left, then right) to the queue.
• Add the current level list to the result.

⏳ Time & Space Complexity

• Time Complexity: $O(N)$. Each node is visited once.
• Space Complexity: $O(N)$ or more specifically $O(W)$ where $W$ is the maximum width of the tree (for the queue).

🚀 Code Implementations

C++

#include <vector>
#include <queue>
#include <iostream>
using namespace std;
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};
class Solution {
public:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> result;
if (!root) return result;
queue<TreeNode*> q;
q.push(root);
while (!q.empty()) {
int size = q.size();
vector<int> currentLevel;
for (int i = 0; i < size; i++) {
TreeNode* node = q.front();
q.pop();
currentLevel.push_back(node->val);
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
}
result.push_back(currentLevel);
}
return result;
}
};

Python

from typing import List, Optional
from collections import deque
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]]:
if not root:
return []
result = []
queue = deque([root])
while queue:
level_size = len(queue)
current_level = []
for _ in range(level_size):
node = queue.popleft()
current_level.append(node.val)
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
result.append(current_level)
return result

Java

import java.util.*;
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<>();
if (root == null) return result;
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> currentLevel = new ArrayList<>();
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
currentLevel.add(node.val);
if (node.left != null) queue.offer(node.left);
if (node.right != null) queue.offer(node.right);
}
result.add(currentLevel);
}
return result;
}
}

🌍 Real-World Analogy

Organizational Chart:

Traversing a company hierarchy level by level. First the CEO (Level 1), then all VPs (Level 2), then all Directors (Level 3), and so on.

🎯 Summary

✅ Queue is Key: BFS strictly relies on a FIFO queue to process nodes level by level.