ZigZag Tree Traversal II in JavaScript (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 visual representations of tree structures.

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

Approach

To solve this problem, we can use a breadth-first search (BFS) approach with a queue to keep track of nodes at each level. We will also use a flag to alternate the direction of traversal for each level.

Here is a step-by-step approach:

Initialize a queue with the root node and a flag to indicate the direction of traversal.
While the queue is not empty, process each level of the tree:
- Determine the number of nodes at the current level.
- For each node at the current level, add its value to a temporary list.
- Add the children of each node to the queue for the next level.
- If the flag indicates a right-to-left traversal, reverse the temporary list.
- Add the temporary list to the result list and toggle the flag.

Algorithm

Here is a detailed breakdown of the algorithm:

Initialize an empty result list and a queue with the root node.
Initialize a boolean flag to indicate the direction of traversal (starting with left-to-right).
While the queue is not empty:
- Determine the number of nodes at the current level.
- Initialize an empty list to store the values of nodes at the current level.
- For each node at the current level:
  - Remove the node from the queue and add its value to the temporary list.
  - Add the node's children to the queue for the next level.
- If the flag indicates a right-to-left traversal, reverse the temporary list.
- Add the temporary list to the result list.
- Toggle the flag for the next level.

Code Implementation


/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */

/**
 * @param {TreeNode} root
 * @return {number[][]}
 */
var zigzagLevelOrder = function(root) {
    if (!root) return [];
    
    const result = [];
    const queue = [root];
    let leftToRight = true;
    
    while (queue.length > 0) {
        const levelSize = queue.length;
        const currentLevel = [];
        
        for (let i = 0; i < levelSize; i++) {
            const node = queue.shift();
            currentLevel.push(node.val);
            
            if (node.left) queue.push(node.left);
            if (node.right) queue.push(node.right);
        }
        
        if (!leftToRight) {
            currentLevel.reverse();
        }
        
        result.push(currentLevel);
        leftToRight = !leftToRight;
    }
    
    return 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 function should return an empty list.
A tree with only one node: The function should return a list with one sublist containing the single node's value.
A tree with multiple levels but some levels having only one node: The function should handle these cases correctly and alternate the direction of traversal.

Testing

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

An empty tree: zigzagLevelOrder(null) should return [].
A tree with one node: zigzagLevelOrder(new TreeNode(1)) should return [[1]].
A tree with multiple levels: Use the example provided in the problem statement.

Use a testing framework like Jest to automate the testing process.

Thinking and Problem-Solving Tips

When approaching such problems, it is helpful to:

Break down the problem into smaller parts and solve each part step-by-step.
Use diagrams to visualize the tree structure and the traversal process.
Practice similar problems to strengthen your understanding of tree traversal algorithms.

Conclusion

In this blog post, we discussed the zigzag level order traversal of a binary tree. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for developing strong problem-solving skills in computer science.

We encourage you to practice and explore further to deepen your understanding.