Zigzag Level Order Traversal in DSA Typescript

Mental Model

We visit nodes level by level, but switch direction each level, going left to right then right to left alternately.

Analogy: Imagine reading lines of text where the first line you read left to right, the next line right to left, then left to right again, like zigzagging across the page.

      1
     / \
    2   3
   / \   \
  4   5   6

Level 0: 1 (left to right)
Level 1: 3 -> 2 (right to left)
Level 2: 4 -> 5 -> 6 (left to right)

Dry Run Walkthrough

Input: Binary tree: 1 as root, 2 and 3 as children of 1, 4 and 5 as children of 2, 6 as right child of 3

Goal: Traverse the tree level by level, but alternate direction each level to get zigzag order

Step 1: Start at root level (level 0), direction left to right

Level 0: [1]
Output: [[1]]

Why: We begin traversal from the root, going left to right

Step 2: Move to level 1, switch direction to right to left

Level 1 nodes: 2, 3
Output: [[1], [3, 2]]

Why: We reverse order at this level to get zigzag pattern

Step 3: Move to level 2, switch direction back to left to right

Level 2 nodes: 4, 5, 6
Output: [[1], [3, 2], [4, 5, 6]]

Why: Direction alternates again to left to right

Result:

Final output: [[1], [3, 2], [4, 5, 6]]

Annotated Code

DSA Typescript

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
    this.val = val === undefined ? 0 : val;
    this.left = left === undefined ? null : left;
    this.right = right === undefined ? null : right;
  }
}

function zigzagLevelOrder(root: TreeNode | null): number[][] {
  if (!root) return [];
  const result: number[][] = [];
  const queue: TreeNode[] = [root];
  let leftToRight = true;

  while (queue.length > 0) {
    const levelSize = queue.length;
    const levelNodes: number[] = [];

    for (let i = 0; i < levelSize; i++) {
      const node = queue.shift()!; // remove front node
      // Add node value based on current direction
      if (leftToRight) {
        levelNodes.push(node.val); // left to right
      } else {
        levelNodes.unshift(node.val); // right to left
      }
      // Add children for next level
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
    result.push(levelNodes);
    leftToRight = !leftToRight; // flip direction
  }
  return result;
}

// Driver code to test
const root = new TreeNode(1,
  new TreeNode(2, new TreeNode(4), new TreeNode(5)),
  new TreeNode(3, null, new TreeNode(6))
);
const output = zigzagLevelOrder(root);
console.log(JSON.stringify(output));

if (!root) return [];

handle empty tree edge case

const queue: TreeNode[] = [root];

initialize queue with root for level order traversal

while (queue.length > 0) {

process nodes level by level until none left

const levelSize = queue.length;

determine number of nodes at current level

const node = queue.shift()!;

remove front node to process current level

if (leftToRight) { levelNodes.push(node.val); } else { levelNodes.unshift(node.val); }

add node value in correct order based on direction

if (node.left) queue.push(node.left); if (node.right) queue.push(node.right);

enqueue children for next level

leftToRight = !leftToRight;

flip direction for next level

Output

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

Complexity Analysis

Time: O(n) because each node is visited exactly once during traversal

Space: O(n) because the queue and result store nodes and values for each level

vs Alternative: Compared to normal level order traversal, zigzag adds a small overhead for reversing order but still remains O(n)

Edge Cases

empty tree (root is null)

returns empty array immediately

DSA Typescript

if (!root) return [];

tree with single node

returns array with one level containing that node

DSA Typescript

const queue: TreeNode[] = [root];

tree with only left or only right children

still traverses levels correctly alternating direction

DSA Typescript

if (node.left) queue.push(node.left); if (node.right) queue.push(node.right);

When to Use This Pattern

When a problem asks for level order traversal but with alternating directions each level, use zigzag level order traversal to handle the direction switch cleanly.

Common Mistakes

Mistake: Always appending node values left to right without reversing on alternate levels
Fix: Use unshift to add values at front on alternate levels to reverse order

Mistake: Reversing the entire level array after collecting all nodes instead of inserting in correct order
Fix: Insert nodes in correct order during traversal to avoid extra reversal step

Summary

Traverses a binary tree level by level, alternating direction each level to create a zigzag pattern.

Use when you need to visit nodes in a breadth-first manner but want to alternate the order of nodes at each level.

The key insight is to switch the order of inserting node values at each level instead of reversing after traversal.