21. ZigZag Tree Traversal

The problem can be found at the following link: Question Link

My Approach

• Initialize an empty vector ans to store the zigzag traversal result.

• Check if the root is null. If so, return the empty ans vector.

• Use a queue q to perform level order traversal.

• Start by pushing the root node into the queue.

• Initialize a boolean variable left to true to control the direction of traversal.

• While the queue is not empty :

  • Get the size of the queue to process nodes at the current level.

  • Create a temporary vector vec to store node values at the current level.

  • Iterate through the nodes at the current level.

    • Dequeue a node from the front of the queue.

    • the index to insert the node value based on the traversal direction.

    • Store the node value at the determined index in vec.

    • If the dequeued node has left or right child, enqueue them.

  • Toggle the value of left to switch the direction for the next level.

  • Append the values stored in vec to the result vector ans.

• Return the result vector ans containing the zigzag traversal of the tree.

Time and Auxiliary Space Complexity

• Time Complexity: The time complexity of this approach is O(N), where N is the number of nodes in the binary tree.

• Auxiliary Space Complexity: The auxiliary space complexity is O(N), where N is the maximum number of nodes at any level of the tree.

Code (C++)

class Solution{
    public:
    //Function to store the zig zag order traversal of tree in a list.
    vector<int>ans;
    vector <int> zigZagTraversal(Node* root)
    {
        if (root==nullptr)
            return ans;
        queue<Node*> q;
        q.push(root);
        bool left=true;
        while (!q.empty())
        {
            int size=q.size();
            vector<int>vec(size);
            for (int i=0;i<size;i++)
            {
                Node* temp=q.front();
                q.pop();
                int ind=left ? i : (size-1-i);
                vec[ind]=temp->data;
                if (temp->left)
                    q.push(temp->left);
                if (temp->right)
                    q.push(temp->right);
            }
            left=!left;
            ans.insert(ans.end(), vec.begin(), vec.end());
        }
        return ans;
    }
};

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