Zigzag Level Order Traversal in DSA C++ - 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.

Specifically, how does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

vector> zigzagLevelOrder(TreeNode* root) {
    vector> result;
    if (!root) return result;
    queue q; q.push(root);
    bool leftToRight = true;
    while (!q.empty()) {
        int size = q.size();
        vector level(size);
        for (int i = 0; i < size; ++i) {
            TreeNode* node = q.front(); q.pop();
            int index = leftToRight ? i : (size - 1 - i);
            level[index] = node->val;
            if (node->left) q.push(node->left);
            if (node->right) q.push(node->right);
        }
        leftToRight = !leftToRight;
        result.push_back(level);
    }
    return result;
}
    

This code visits each node level by level, alternating the order of values collected to create a zigzag pattern.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once in the tree.
• How many times: Exactly once per node, inside the while loop and for loop combined.

How Execution Grows With Input

As the number of nodes (n) increases, the code visits each node once, so the work grows directly with n.

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

Pattern observation: The work grows linearly as the tree size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows directly in proportion to 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 values collected, but each node is still visited once, so the time stays linear.

Interview Connect

Understanding this traversal helps you show you can handle tree traversals with order changes, a common interview skill.

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 does Zigzag Level Order Traversal of a binary tree mean?

easy

A. Traverse nodes level by level, alternating left to right and right to left

B. Traverse nodes only from left to right at each level

C. Traverse nodes only from right to left at each level

D. Traverse nodes randomly without any order

Solution

1. Step 1: Understand level order traversal

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

2. Step 2: Add zigzag pattern

   Zigzag means alternate direction each level: first left to right, then right to left, and so on.

3. Final Answer:

   Traverse nodes level by level, alternating left to right and right to left -> Option A

4. Quick Check:

   Zigzag = alternate directions [OK]

Hint: Remember zigzag means alternate directions each level [OK]

Common Mistakes:

• Thinking traversal is always left to right
• Confusing zigzag with preorder or inorder traversal
• Ignoring the alternating direction pattern

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

easy

A. std::list q; q.push_front(root);

B. std::stack q; q.push(root);

C. std::vector q; q.push_back(root);

D. std::queue q; q.push(root);

Solution

1. Step 1: Identify correct data structure for level order

   Level order traversal uses a queue to process nodes FIFO (first in, first out).

2. Step 2: Check code snippets

   std::queue q; q.push(root); uses std::queue and pushes root correctly. Others use stack, vector, or list incorrectly for this purpose.

3. Final Answer:

   std::queue q; q.push(root); -> Option D

4. Quick Check:

   Queue for level order = std::queue q; q.push(root); [OK]

Hint: Use queue for level order traversal, not stack or vector [OK]

Common Mistakes:

• Using stack instead of queue
• Using vector push_back but no FIFO behavior
• Using list with push_front which reverses order

3. What is the output of the zigzag level order traversal for this tree?

    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 has node 1, so output is [1].

2. Step 2: Traverse level 1 right to left

   Level 1 nodes are 2 and 3. Zigzag means reverse order here: [3, 2].

3. Step 3: Traverse level 2 left to right

   Level 2 nodes are 4, 5, 6 from left to right: [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 order on alternate levels
• Mixing node order at last level
• Confusing left and right child order

4. Identify the bug in this C++ snippet for zigzag traversal:

std::vector zigzagLevelOrder(TreeNode* root) {
  std::vector result;
  if (!root) return result;
  std::queue q;
  q.push(root);
  bool leftToRight = true;
  while (!q.empty()) {
    int size = q.size();
    std::vector<int> level(size);
    for (int i = 0; i < size; ++i) {
      TreeNode* node = q.front(); q.pop();
      int index = leftToRight ? i : size - i - 1;
      level[index] = node->val;
      if (node->left) q.push(node->left);
      if (node->right) q.push(node->right);
    }
    result.push_back(level);
  }
  return result;
}

medium

A. Using vector<int> level(size) instead of vector<int> level

B. Incorrect queue initialization

C. Missing toggle of leftToRight after each level

D. Not checking if root is null

Solution

1. Step 1: Analyze loop for toggling direction

   The variable leftToRight controls direction but is never changed inside the loop.

2. Step 2: Identify missing toggle

   Without toggling leftToRight = !leftToRight; after processing each level, zigzag pattern won't alternate.

3. Final Answer:

   Missing toggle of leftToRight after each level -> Option C

4. Quick Check:

   Toggle direction each level [OK]

Hint: Toggle direction flag after each level to alternate traversal [OK]

Common Mistakes:

• Forgetting to flip leftToRight boolean
• Incorrect indexing in level vector
• Not pushing child nodes properly

5. Given a binary tree, how would you modify the zigzag level order traversal to return a flat vector of node values instead of a vector of vectors? For example, for tree:

    1
   / \
  2   3
 /     \
4       5

The output should be: [1, 3, 2, 4, 5]

hard

A. Use a stack instead of queue to store nodes for zigzag traversal

B. Perform normal zigzag traversal, then flatten the result vectors into one vector

C. Traverse all nodes left to right ignoring zigzag pattern

D. Return only the last level nodes in zigzag order

Solution

1. Step 1: Perform standard zigzag traversal

   Use queue and toggle direction each level to get vector of vectors with zigzag order.

2. Step 2: Flatten the vector of vectors

   Combine all inner vectors into a single vector preserving order to get flat list.

3. Final Answer:

   Perform normal zigzag traversal, then flatten the result vectors into one vector -> Option B

4. Quick Check:

   Flatten zigzag levels to one vector [OK]

Hint: Flatten after zigzag traversal to get single list [OK]

Common Mistakes:

• Trying to zigzag directly into flat vector without levels
• Using stack incorrectly for zigzag
• Ignoring zigzag pattern and traversing left to right only