Level order Traversal / Level order traversal in spiral form

Dec 18, 2024 • 5 mins read

Problem Overview

One common problem encountered when working with binary trees is the need to traverse the tree in specific patterns. The problem "Binary Tree Level Order Traversal" seeks to address this by requiring the traversal of a binary tree level by level, starting from the root and proceeding to each subsequent level in order. This is also known as Breadth-First Search (BFS) traversal of a binary tree.

Understanding the Problem

Given a binary tree, we need to return the level order traversal of its nodes' values. Each level's values should be returned as separate lists, encapsulated within a larger list. This means the output should be structured as a list of lists where each inner list contains the values of the nodes at a particular level.

Example

Consider the binary tree shown below:

The level order traversal for this tree should return:

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

Key DSA Concepts and Theory

Binary Tree

A binary tree is a hierarchical structure in which each node has at most two children, referred to as the left child and the right child. It is widely used in data structures, algorithms, and their applications due to its recursive nature that allows for effective traversal and manipulation.

Breadth-First Search (BFS)

Breadth-First Search is an algorithm to traverse or search tree or graph data structures. It starts at the root (selecting some arbitrary node as the root in the case of graph) and explores all the neighbor nodes at the present depth prior to moving on to the nodes at the next depth level.

In the context of binary trees, BFS equates to visiting each level fully before proceeding to the next one. This approach naturally maps to level order traversal.

Solution Approach

To achieve the desired level order traversal, we can make use of a queue data structure. The essence of this approach is to repeatedly process nodes level by level, each time enqueuing the children of nodes currently being processed, until all nodes in the binary tree have been traversed.

Here's the detailed step-by-step breakdown of the algorithm:

Initialize a Queue: Start by initializing a queue and adding the root node to it. Additionally, prepare a list (result) to store the final level order traversal.
Perform Level Order Traversal:
- While the queue is not empty, perform the following operations:
  - Determine the number of nodes at the current level (level_size).
  - Initialize an empty list to temporarily hold the values for the current level.
  - For each node at the current level, dequeue the node and add its value to the temporary list.
  - If left or right children of the current node exist, enqueue them for future processing.
- Append the temporary list of level values to the result.
Return the Result: Once the traversal is complete, return the result which contains lists of values corresponding to each level.

C++ Code Implementation

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

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 level_size = q.size();
        vector<int> level;
        
        for (int i = 0; i < level_size; ++i) {
            TreeNode* current = q.front();
            q.pop();
            level.push_back(current->val);
            
            if (current->left) q.push(current->left);
            if (current->right) q.push(current->right);
        }
        
        result.push_back(level);
    }
    
    return result;
}

Java Code Implementation

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int x) { val = x; }
}

public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) return result;
        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {
            List<Integer> level = new ArrayList<>();
            int level_size = queue.size();
            
            for (int i = 0; i < level_size; i++) {
                TreeNode current = queue.poll();
                level.add(current.val);
                
                if (current.left != null) queue.offer(current.left);
                if (current.right != null) queue.offer(current.right);
            }
            
            result.add(level);
        }
        
        return result;
    }
}

Time and Space Complexity Analysis

Time Complexity: The time complexity is O(n), where n is the number of nodes in the binary tree. This is because each node is processed exactly once.
Space Complexity: The space complexity is O(n) in the worst case, which occurs when the binary tree is completely unbalanced (like a linked list) where the queue contains n/2 nodes at one point. In a more balanced tree, the space complexity will still be bound by the largest number of nodes in any level, also approaching O(n) in the worst-case scenario.

Common Mistakes to Avoid

Null Root Node: Always check for a null root node before commencing the traversal. Failing to do so will result in null pointer exceptions.
Queue Management: Ensure that nodes are enqueued and dequeued properly. Mismanagement of the queue can lead to infinite loops or skipping of nodes.
Correct Node Order: Be wary of the order of enqueuing left and right children as changing the order can lead to incorrect traversal results.

Additional Resources and References

This problem illustrates the application of BFS in dealing with hierarchical data structures and is a fundamental technique in tree processing tasks.

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