Zigzag Level Order Traversal in DSA Java - Time & Space Complexity

Time Complexity: Zigzag Level Order Traversal

O(n)

Understanding Time Complexity

We want to understand how the time needed to traverse a tree in zigzag order changes as the tree grows.

How does the number of steps grow when the tree has more nodes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public List> zigzagLevelOrder(TreeNode root) {
    List> result = new ArrayList<>();
    if (root == null) return result;
    Queue queue = new LinkedList<>();
    queue.offer(root);
    boolean leftToRight = true;
    while (!queue.isEmpty()) {
        int size = queue.size();
        LinkedList level = new LinkedList<>();
        for (int i = 0; i < size; i++) {
            TreeNode node = queue.poll();
            if (leftToRight) level.addLast(node.val);
            else level.addFirst(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;
}
    

This code visits each node of a binary tree level by level, alternating the order of values collected at each level.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once in the while loop and inner for loop.
• How many times: Each node is processed exactly one time.

How Execution Grows With Input

As the number of nodes increases, the steps increase roughly the same amount.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The work grows linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the traversal grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Because of the zigzag pattern, the traversal takes more than linear time."

[OK] Correct: The zigzag only changes the order of adding values, but each node is still visited once, so time grows linearly.

Interview Connect

Understanding this traversal helps you show you can handle tree traversals with variations, a common skill interviewers look for.

Self-Check

"What if we used a depth-first search instead of breadth-first search for zigzag traversal? How would the time complexity change?"

Practice

1. What is the main difference between a normal level order traversal and a zigzag level order traversal of a binary tree?

easy

A. Zigzag traversal alternates the direction of node visits at each level.

B. Zigzag traversal visits nodes only on odd levels.

C. Zigzag traversal visits nodes in reverse order at all levels.

D. Zigzag traversal visits nodes randomly without any order.

Solution

1. Step 1: Understand normal level order traversal

   Normal level order traversal visits nodes level by level from left to right.

2. Step 2: Understand zigzag traversal difference

   Zigzag traversal changes direction at each level, alternating between left-to-right and right-to-left.

3. Final Answer:

   Zigzag traversal alternates the direction of node visits at each level. -> Option A

4. Quick Check:

   Zigzag = alternate direction [OK]

Hint: Remember zigzag flips direction every level [OK]

Common Mistakes:

• Thinking zigzag visits only odd levels
• Assuming zigzag always goes right to left
• Confusing zigzag with random order

2. Which of the following Java code snippets correctly initializes a queue for level order traversal?

easy

A. Queue<TreeNode> queue = new ArrayList<>();

B. Queue<TreeNode> queue = new HashSet<>();

C. Queue<TreeNode> queue = new Stack<>();

D. Queue<TreeNode> queue = new LinkedList<>();

Solution

1. Step 1: Identify correct queue implementation

   Java's Queue interface is commonly implemented by LinkedList for FIFO behavior.

2. Step 2: Check other options

   ArrayList is a list, not a queue; Stack is LIFO, not queue; HashSet is a set, no order.

3. Final Answer:

   Queue<TreeNode> queue = new LinkedList<>(); -> Option D

4. Quick Check:

   Queue = LinkedList [OK]

Hint: Use LinkedList for queue in Java [OK]

Common Mistakes:

• Using ArrayList instead of Queue
• Using Stack which is LIFO, not FIFO
• Using HashSet which has no order

3. Given the binary tree below, what is the zigzag level order traversal output?

    1
   / \
  2   3
 / \   \
4   5   6

medium

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

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

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

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

Solution

1. Step 1: Traverse level 0 left to right

   Level 0 nodes: [1]

2. Step 2: Traverse level 1 right to left

   Level 1 nodes: [2, 3] reversed to [3, 2]

3. Step 3: Traverse level 2 left to right

   Level 2 nodes: [4, 5, 6]

4. Final Answer:

   [[1], [3, 2], [4, 5, 6]] -> Option A

5. Quick Check:

   Zigzag levels alternate direction [OK]

Hint: Alternate direction each level starting left to right [OK]

Common Mistakes:

• Not reversing nodes on odd levels
• Reversing nodes on even levels incorrectly
• Mixing node order at last level

4. Consider the following Java code snippet for zigzag traversal. What is the error?

Queue<TreeNode> queue = new LinkedList<>();
boolean leftToRight = true;
while (!queue.isEmpty()) {
    int size = queue.size();
    List<Integer> level = new ArrayList<>();
    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.add(node.left);
        if (node.right != null) queue.add(node.right);
    }
    leftToRight = !leftToRight;
}

medium

A. The boolean flag leftToRight should not be toggled.

B. The queue should be a Stack for zigzag traversal.

C. The for loop condition should be i < size, not i <= size.

D. The level list should be a LinkedList instead of ArrayList.

Solution

1. Step 1: Analyze for loop condition

   The loop runs from i = 0 to i <= size, which means it runs size+1 times, causing poll() to return null once.

2. Step 2: Correct loop condition

   The loop should run exactly size times, so condition must be i < size.

3. Final Answer:

   The for loop condition should be i < size, not i <= size. -> Option C

4. Quick Check:

   Loop runs size times [OK]

Hint: Use i < size for loops over queue size [OK]

Common Mistakes:

• Using i <= size causing extra iteration
• Using Stack instead of Queue
• Not toggling direction flag

5. You want to implement zigzag level order traversal but optimize space by using a single deque (double-ended queue). Which approach correctly uses the deque to achieve zigzag traversal?

hard

A. Use deque to add nodes at front for left-to-right levels and remove from end; add nodes at end for right-to-left levels and remove from front.

B. Use deque to add nodes at the end for left-to-right levels and remove from front; add nodes at front for right-to-left levels and remove from end.

C. Use deque to add nodes only at end and remove from front for all levels.

D. Use deque to add nodes only at front and remove from end for all levels.

Solution

1. Step 1: Understand deque usage for zigzag

   For left-to-right levels, add children at the end and remove nodes from front to process in order.

2. Step 2: Reverse direction for right-to-left levels

   For right-to-left levels, add children at front and remove nodes from end to process in reverse order.

3. Final Answer:

   Use deque to add nodes at the end for left-to-right levels and remove from front; add nodes at front for right-to-left levels and remove from end. -> Option B

4. Quick Check:

   Deque add/remove ends alternate by level [OK]

Hint: Alternate deque ends for add/remove by level [OK]

Common Mistakes:

• Adding/removing from same end always
• Not reversing add/remove ends per level
• Using only queue or stack instead of deque