Unlock 10+ LeetCode Problems with Level Order Traversal Template

LeetCode can feel overwhelming at times, but I’ve found a method that really works for me: focusing on problem patterns. By solving multiple problems within the same pattern and making only minor changes to my approach, I’ve been able to understand the core concepts much better. It’s a simple but powerful shift that’s made all the difference in my preparation, and I hope it can help others too.

Understanding Level Order Traversal

Level order traversal is a method for exploring binary trees where we visit nodes level by level, from left to right. This approach is surprisingly versatile and forms the basis for solving numerous tree-related problems.

Here’s a basic implementation of level order traversal:

Problem 1 : Level Order Traversal

public List<List<Integer>> levelOrder(TreeNode root) {
    Map<Integer, List<Integer>> level = new HashMap<>();
    levelOrder(root, 0, level);
    return new ArrayList<>(level.values());
}

private void levelOrder(TreeNode root, int currLevel, Map<Integer, List<Integer>> level) {
    if (root == null) return;
    level.putIfAbsent(currLevel, new ArrayList<>());
    level.get(currLevel).add(root.val);
    levelOrder(root.left, currLevel + 1, level);
    levelOrder(root.right, currLevel + 1, level);
}

Let’s break down this implementation:

1. We use a HashMap to store nodes at each level. The key is the level number, and the value is a list of node values at that level.
2. We start at the root (level 0) and work our way down the tree.
3. For each node, we add its value to the list corresponding to its level in the HashMap.
4. We recursively process the left and right children, incrementing the level.

This basic pattern can be adapted to solve a wide range of problems.

Applying the Pattern

Problem 2: The “Binary Tree Level Order Traversal II” problem requires us to return the levels from bottom to top. We can solve this by simply reversing our result:

public List<List<Integer>> levelOrderBottom(TreeNode root) {
    Map<Integer, List<Integer>> level = new HashMap<>();
    levelOrder(root, 0, level);
    List<List<Integer>> result = new ArrayList<>(level.values());
    return result.reversed();
}
//Use the same level order traversal method

Problem 3 : For the zigzag traversal, we alternate the direction of traversal for each level

public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
    Map<Integer, List<Integer>> level = new HashMap<>();
    levelOrder(root, 0, level);
    List<List<Integer>> result = new ArrayList<>();
    int maxLevel = level.size();
    for (int i = 0; i < maxLevel; i++) {
        List<Integer> currentLevel = level.get(i);
        if (i % 2 == 1) {
            Collections.reverse(currentLevel);
        }
        result.add(currentLevel);
    }
    return result;
}
//Use the same level order traversal method

Expanding Your Problem-Solving Toolkit

Once you’re comfortable with this pattern, you can adapt it to solve various other problems

Problem 4 : Maximum Level Sum of a Binary Tree: Keep track of the sum at each level and find the maximum.

Problem 5 :Binary Tree Right Side View: Take the last element from each level.

Problem 6:Average of Levels in Binary Tree: Calculate the average of node values at each level.

Problem 7: Find the largest value in each row : Modify the template to store only the maximum value for each level.

Problem 8 : Even Odd Tree : Verify value conditions for each level during traversal.

Problem 9 : Check for completeness of the binary tree : Use level order traversal and check for gaps between nodes.

Problem 10 : N-ary tree level order traversal : Adapt the template to handle multiple children instead of just left and right.

By mastering level order traversal and practicing these problems, you’ll significantly enhance your ability to solve tree-related coding challenges. This focused approach can save you time and boost your confidence in technical interviews.

Remember, the key to success is not just solving problems, but understanding the underlying patterns. Happy coding!