Level Order Tree Traversal in C++ with Time Complexity Analysis

Given a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

Example:

Input: [3, 9, 20, null, null, 15, 7]
    3
   / \
  9  20
    /  \
   15   7

Output: 
[
  [3],
  [9, 20],
  [15, 7]
]

Understanding the Problem

The core challenge of this problem is to traverse a binary tree level by level, from left to right. This type of traversal is known as level order traversal. It is significant in many applications such as breadth-first search (BFS) in graphs, serialization/deserialization of trees, and more.

Potential pitfalls include not handling null nodes correctly and not maintaining the order of nodes at each level.

Approach

To solve this problem, we can use a queue data structure to help us perform a breadth-first search (BFS) on the tree. The queue will help us keep track of nodes at the current level and their children for the next level.

Naive Solution

A naive solution might involve using a recursive approach to traverse each level, but this can be inefficient and complex to implement.

Optimized Solution

The optimized solution involves using a queue to perform BFS. This approach is efficient and straightforward:

Initialize a queue and push the root node into it.
While the queue is not empty, process each node level by level.
For each node, push its children into the queue.
Collect the values of nodes at each level and store them in a result list.

Algorithm

Here is a step-by-step breakdown of the BFS algorithm for level order traversal:

Initialize an empty queue and push the root node into it.
While the queue is not empty, do the following:
- Initialize an empty list to store the values of nodes at the current level.
- Get the number of nodes at the current level (size of the queue).
- For each node at the current level, do the following:
  - Pop the node from the queue.
  - Add its value to the current level list.
  - Push its left and right children (if they exist) into the queue.
- Add the current level list to the result list.

Code Implementation

#include <iostream>
#include <vector>
#include <queue>

using namespace std;

// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

vector<vector<int>> levelOrder(TreeNode* root) {
    vector<vector<int>> result;
    if (!root) return result;

    queue<TreeNode*> q;
    q.push(root);

    while (!q.empty()) {
        int levelSize = q.size();
        vector<int> currentLevel;

        for (int i = 0; i < levelSize; ++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);
        }

        result.push_back(currentLevel);
    }

    return result;
}

// Helper function to create a new tree node
TreeNode* newNode(int data) {
    TreeNode* node = new TreeNode(data);
    node->left = node->right = NULL;
    return node;
}

// Main function to test the level order traversal
int main() {
    TreeNode* root = newNode(3);
    root->left = newNode(9);
    root->right = newNode(20);
    root->right->left = newNode(15);
    root->right->right = newNode(7);

    vector<vector<int>> result = levelOrder(root);

    for (const auto& level : result) {
        for (int val : level) {
            cout << val << " ";
        }
        cout << endl;
    }

    return 0;
}

Complexity Analysis

The time complexity of the BFS approach is O(n), where n is the number of nodes in the tree. This is because each node is processed exactly once. The space complexity is also O(n) due to the queue used to store nodes at each level.

Edge Cases

Potential edge cases include:

An empty tree (root is null).
A tree with only one node.
A tree where all nodes are on one side (left or right skewed).

Each of these cases is handled by the BFS approach as it processes nodes level by level and checks for null nodes.

Testing

To test the solution comprehensively, consider the following test cases:

An empty tree.
A tree with a single node.
A balanced tree.
A left-skewed tree.
A right-skewed tree.

Using a variety of test cases ensures that the solution handles all possible scenarios.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Understand the problem requirements and constraints thoroughly.
Think about different traversal methods and their applications.
Use data structures like queues or stacks to help with traversal.
Break down the problem into smaller steps and solve each step methodically.

Conclusion

Level order traversal is a fundamental problem in tree data structures. Understanding and implementing this traversal helps in solving more complex tree-related problems. Practice and familiarity with different traversal methods are key to mastering tree algorithms.

Additional Resources

For further reading and practice, consider the following resources: