LeetCode 515: Find Largest Value in Each Tree Row | Easy Explanantion | C++ | CPP

Binary Trees are a foundational data structure in computer science, and solving problems like LeetCode 515: Find Largest Value in Each Tree Row helps us understand the core principles of traversal and optimization. In this blog post, we will break down the problem, explain the approaches to solve it, analyze their time and space complexities, and highlight important concepts such as BFS, DFS, and edge case handling.

Problem Statement

Given the root of a binary tree, return an array of the largest value in each row of the tree (0-indexed).

Example 1:

Input: root = [1,3,2,5,3,null,9]

Output: [1, 3, 9]




Example 2:

Input: root = [1,2,3]

    1
   /  \
  2   3

Output: [1, 3]




Constraints:

• The number of nodes in the tree will be in the range [0, 10^4].

To solve this problem, we need to traverse the binary tree level by level and find the maximum value in each row. Below are the approaches, from brute force to optimized solutions:

1. Brute Force Approach (Recursive DFS)

In this approach, we use Depth First Search (DFS) to traverse the tree and maintain a map of row indices to their maximum values. For each node, we update the maximum value for its respective row.

Algorithm:

1. Use a helper function to traverse the tree recursively.

2. Keep track of the current depth (row index).

3. Update the maximum value for each row during the traversal.

4. At the end, collect the maximum values for each row.

Code:

class Solution

{

public:

    void dfs(TreeNode *node, int depth, vector<int> &maxValues)

    {

        if (!node)

            return;

        if (depth == maxValues.size())

        {

            maxValues.push_back(node->val);

        }

        else

        {

            maxValues[depth] = max(maxValues[depth], node->val);

        }

        dfs(node->left, depth + 1, maxValues);

        dfs(node->right, depth + 1, maxValues);

    }

    vector<int> largestValues(TreeNode *root)

    {

        vector<int> maxValues;

        dfs(root, 0, maxValues);

        return maxValues;

    }

};

Time Complexity:

• Traverses all n nodes: O(n).

Space Complexity:

• Recursion stack depth for skewed trees: O(h), where h is the height of the tree.

2. Optimized BFS Approach (Level Order Traversal)

The most efficient way to solve this problem is using Breadth First Search (BFS). This method leverages a queue to perform a level-order traversal of the tree, ensuring we process all nodes in a row before moving to the next.

Algorithm:

1. Initialize a queue and push the root node.

2. For each level, iterate through all nodes, keeping track of the maximum value.

3. At the end of each level, store the maximum value in the result.

Code:

class Solution

{

public:

    vector<int> largestValues(TreeNode *root)

    {

        vector<int> ans;

        if (!root)

            return ans;

        queue<TreeNode *> q;

        q.push(root);

        while (!q.empty())

        {

            int levelSize = q.size();

            int maxVal = INT_MIN;

            for (int i = 0; i < levelSize; ++i)

            {

                TreeNode *node = q.front();

                q.pop();

                maxVal = max(maxVal, node->val);

                if (node->left)

                    q.push(node->left);

                if (node->right)

                    q.push(node->right);

            }

            ans.push_back(maxVal);

        }

        return ans;

    }

};

Explanation:

• The queue ensures we process nodes level by level.

• levelSize helps isolate each row's nodes.

• At the end of each level, we record the largest value.

Time Complexity:

• Processing all nodes: O(n).

Space Complexity:

• Queue size for the largest level: O(w), where w is the maximum width of the tree.

Edge Cases

1. Empty Tree: If the root is NULL, return an empty array.

2. Single Node Tree: Return an array containing the root's value.

3. Negative Values: Ensure the solution works when all node values are negative.

Optimization Tips

• Avoid unnecessary NULL checks during traversal by initializing queues and recursive calls properly.

• Use iterative methods like BFS for better control over memory usage in deep or skewed trees.

The problem "Find Largest Value in Each Tree Row" highlights the importance of traversal techniques in solving binary tree problems. The BFS approach is generally the most efficient for this task due to its level-by-level processing.

By understanding the nuances of BFS and DFS, you can extend these techniques to solve more complex tree-based problems. Happy coding!

Tags

• Binary Tree
  Breadth First Search
  Depth First Search
  Tree Traversal
  Level Order Traversal
  LeetCode Solutions
  C++ Programming
  Algorithm Optimization
  Coding Interview Questions