ZigZag Tree Traversal in Java (Time Complexity: O(n))

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

Example:

Input: root = [3, 9, 20, null, null, 15, 7]
    3
   / \
  9  20
    /  \
   15   7

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

Understanding the Problem

The core challenge of this problem is to traverse the binary tree in a zigzag manner. This means that for each level of the tree, we alternate the direction of traversal. The first level is traversed from left to right, the second from right to left, the third from left to right, and so on.

This type of traversal is significant in scenarios where the order of processing nodes alternates, such as in certain search algorithms or data processing tasks.

Potential pitfalls include not correctly alternating the traversal direction or not handling null nodes properly.

Approach

To solve this problem, we can use a breadth-first search (BFS) approach with a queue to traverse the tree level by level. We will also use a flag to keep track of the current direction of traversal (left-to-right or right-to-left).

Here is a step-by-step approach:

Initialize a queue and add the root node to it.
Use a flag to keep track of the current direction of traversal.
For each level, process all nodes in the queue and add their children to the queue for the next level.
Depending on the current direction, add the node values to a list in the appropriate order.
Toggle the direction flag after processing each level.

Algorithm

Let's break down the algorithm step by step:

Initialize a queue and add the root node to it.
Initialize a flag to keep track of the current direction (left-to-right).
While the queue is not empty, do the following:
- Initialize an empty list to store the current level's node values.
- Get the number of nodes at the current level.
- For each node at the current level, do the following:
  - Remove the node from the queue.
  - Add the node's value to the list in the appropriate order based on the current direction.
  - Add the node's children to the queue for the next level.
- Toggle the direction flag.
- Add the list of node values to the result list.

Code Implementation

import java.util.*;

public class ZigZagTraversal {
    // TreeNode class definition
    public static class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;
        TreeNode(int x) { val = x; }
    }

    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) return result;

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        boolean leftToRight = true;

        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> level = new ArrayList<>(size);

            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                if (leftToRight) {
                    level.add(node.val);
                } else {
                    level.add(0, node.val);
                }

                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }

            result.add(level);
            leftToRight = !leftToRight;
        }

        return result;
    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(3);
        root.left = new TreeNode(9);
        root.right = new TreeNode(20);
        root.right.left = new TreeNode(15);
        root.right.right = new TreeNode(7);

        ZigZagTraversal solution = new ZigZagTraversal();
        List<List<Integer>> result = solution.zigzagLevelOrder(root);
        System.out.println(result);
    }
}

Complexity Analysis

The time complexity of this approach is O(n), where n is the number of nodes in the tree. This is because we visit each node exactly once.

The space complexity is also O(n) due to the queue used for BFS, which in the worst case can hold all the nodes at the deepest level of the tree.

Edge Cases

Some potential edge cases include:

An empty tree (root is null). The expected output is an empty list.
A tree with only one node. The expected output is a list containing one list with that single node's value.
A tree where all nodes are either to the left or right. The algorithm should still correctly alternate the traversal direction.

Testing

To test the solution comprehensively, consider the following test cases:

An empty tree.
A tree with only one node.
A balanced tree with multiple levels.
A tree where all nodes are on one side (left or right).

Using a testing framework like JUnit can help automate and validate these test cases.

Thinking and Problem-Solving Tips

When approaching such problems, it's helpful to:

Break down the problem into smaller, manageable parts.
Consider different traversal methods (BFS, DFS) and choose the one that fits the problem requirements.
Use auxiliary data structures (like queues or stacks) to help manage the traversal state.
Write and test small pieces of code incrementally to ensure correctness.

Conclusion

Understanding and solving the ZigZag Tree Traversal problem helps improve your skills in tree traversal techniques and problem-solving strategies. Practice with similar problems to reinforce these concepts and become proficient in handling tree-based algorithms.

Additional Resources

For further reading and practice, consider the following resources: