Zigzag Level Order Traversal in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the result vector for storing levels.

DSA C++

vector<vector<int>> [1];

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Aresult

Blevels

Coutput

Dzigzag

Attempts:

3 left

💡 Hint

Common Mistakes

Using a single vector instead of vector>

Naming the variable something unrelated

The vector named 'result' is commonly used to store the final zigzag level order traversal.

2fill in blank

medium

Complete the code to check if the current level should be reversed.

DSA C++

if (level % 2 == [1]) { reverse(currentLevel.begin(), currentLevel.end()); }

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A0

B1

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Reversing even levels instead of odd

Using incorrect modulus value

Odd levels (level % 2 == 1) are reversed to achieve zigzag order.

3fill in blank

hard

Fix the error in the queue push operation for left and right children.

DSA C++

if (node->left) q.push([1]);
if (node->right) q.push([2]);

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Anode->left

Bnode->right

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Pushing string literals instead of node pointers

Swapping left and right children

We must push the actual child nodes 'node->left' and 'node->right' into the queue.

4fill in blank

hard

Fill both blanks to correctly iterate over the current level size and pop nodes.

DSA C++

int size = q.[1]();
for (int i = 0; i < size; i++) {
    TreeNode* node = q.[2]();
    q.pop();
    // process node
}

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Asize

Bfront

Cback

Dempty

Attempts:

3 left

💡 Hint

Common Mistakes

Using back() instead of front()

Using empty() instead of size()

Use q.size() to get current level size and q.front() to access the first node in the queue.

5fill in blank

hard

Fill all three blanks to complete the zigzag level order traversal function.

DSA C++

vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
    vector<vector<int>> [1];
    if (!root) return [2];
    queue<TreeNode*> q;
    q.push(root);
    int level = 0;
    while (!q.empty()) {
        int size = q.size();
        vector<int> currentLevel;
        for (int i = 0; i < size; i++) {
            TreeNode* node = q.front();
            q.pop();
            currentLevel.push_back(node->val);
            if (node->left) q.push(node->left);
            if (node->right) q.push(node->right);
        }
        if (level % 2 == 1) {
            reverse(currentLevel.begin(), currentLevel.end());
        }
        [1].push_back(currentLevel);
        level++;
    }
    return [3];
}

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Aresult

Blevels

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names for declaration and return

Returning an uninitialized vector

The vector 'result' is declared, returned, and used to store levels.