Day 1 - Level order traversal.md

Difficulty	Source	Tags
Easy	160 Days of Problem Solving	Tree

🚀 Day 1. Level order traversal 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

Given the root of a binary tree with n nodes, the task is to find its level order traversal.
Level order traversal of a tree is breadth-first traversal for the tree, where we visit nodes level by level.

🔍 Example Walkthrough:

Example 1

Input:

root[] = [1, 2, 3]

Output:

[[1], [2, 3]]

Example 2

Input:

root[] = [10, 20, 30, 40, 50]

Output:

[[10], [20, 30], [40, 50]]

Example 3

Input:

root[] = [1, 3, 2, N, N, N, 4, 6, 5]

Output:

[[1], [3, 2], [4], [6, 5]]

Constraints

• 1 ≤ number of nodes ≤ $10^5$
• 0 ≤ node->data ≤ $10^9$

🎯 My Approach:

1. Use a queue to traverse the tree level by level.
2. Start by pushing the root node into the queue.
3. Process each level:
   • Store the number of nodes in the current level.
   • Traverse all nodes of the level, adding them to the result.
   • Push their left and right children (if they exist) into the queue.
4. Continue the process until all nodes are visited.

This approach ensures that each node is visited exactly once, making it efficient and optimal for level-order traversal.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n), where n is the number of nodes in the tree. Each node is visited exactly once.
• Expected Auxiliary Space Complexity: O(n), since, in the worst case, we store all nodes in the queue.

📝 Solution Code

Code (C++)

class Solution {
public:
    vector<vector<int>> levelOrder(Node *root) {
        if (!root) return {};
        queue<Node *> q({root});
        vector<vector<int>> res;
        while (!q.empty()) {
            res.push_back({});
            for (int i = q.size(); i > 0; i--) {
                Node *n = q.front(); q.pop();
                res.back().push_back(n->data);
                if (n->left) q.push(n->left);
                if (n->right) q.push(n->right);
            }
        }
        return res;
    }
};

🌲 Alternative Approaches

1️⃣ Iterative BFS (Optimized) – Using Queue

This is the most commonly used approach, leveraging BFS (Breadth-First Search) using a queue.

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        if (!root) return {};
        queue<Node*> q({root});
        vector<vector<int>> res;
        while (!q.empty()) {
            vector<int> level;
            for (int i = q.size(); i > 0; i--) {
                auto n = q.front(); q.pop();
                level.push_back(n->data);
                if (n->left) q.push(n->left);
                if (n->right) q.push(n->right);
            }
            res.push_back(move(level));
        }
        return res;
    }
};

2️⃣ Recursive DFS (Depth First Search)

This approach utilizes DFS recursion to store nodes level-wise.

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        vector<vector<int>> res;
        dfs(root, 0, res);
        return res;
    }
private:
    void dfs(Node* root, int lvl, vector<vector<int>>& res) {
        if (!root) return;
        if (lvl == res.size()) res.push_back({});
        res[lvl].push_back(root->data);
        dfs(root->left, lvl + 1, res);
        dfs(root->right, lvl + 1, res);
    }
};

3️⃣ BFS Using Single Loop (Memory Efficient)

This is a slightly more optimized BFS version that avoids extra memory operations.

class Solution {
public:
    vector<vector<int>> levelOrder(Node* root) {
        if (!root) return {};
        queue<Node*> q;
        q.push(root);
        vector<vector<int>> res;
        while (!q.empty()) {
            vector<int> level;
            int n = q.size();
            while (n--) {
                auto node = q.front(); q.pop();
                level.push_back(node->data);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            res.push_back(move(level));
        }
        return res;
    }
};

Comparison of Approaches

Approach	Time Complexity	Space Complexity	Best For
Iterative BFS (Queue) (1️⃣)	O(n)	O(n) (queue storage)	General case (most used)
Recursive DFS (2️⃣)	O(n)	O(n) (recursion stack)	Balanced trees (elegant)
Memory Efficient BFS (3️⃣)	O(n)	O(n) (optimized queue)	Space-efficient traversal

Final Recommendation

• For General Use (Fast & Simple) → Use Iterative BFS (1️⃣)
• For Elegant Recursive Solutions → Use DFS Recursion (2️⃣)
• For Space Efficiency → Use Memory-Efficient BFS (3️⃣)

🚀 The most optimized and commonly used approach is 1️⃣ (Iterative BFS with Queue).

Code (Java)

class Solution {
    public ArrayList<ArrayList<Integer>> levelOrder(Node root) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<>();
        if (root == null) return res;
        Queue<Node> q = new LinkedList<>();
        q.add(root);
        while (!q.isEmpty()) {
            ArrayList<Integer> level = new ArrayList<>();
            for (int i = q.size(); i > 0; i--) {
                Node n = q.poll();
                level.add(n.data);
                if (n.left != null) q.add(n.left);
                if (n.right != null) q.add(n.right);
            }
            res.add(level);
        }
        return res;
    }
}

Code (Python)

class Solution:
    def levelOrder(self, root):
        if not root: return []
        res, q = [], [root]
        while q:
            res.append([n.data for n in q])
            q = [c for n in q for c in (n.left, n.right) if c]
        return res

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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