Level Order (Breadth First Search) Traversal of Binary Tree

Introduction to level order traversal

In the DFS traversal of a binary tree, we access nodes in three different orders: preorder, postorder, and inorder. Now there is another traversal that can access nodes in level-by-level order. This is called level order traversal or breadth-first search traversal. In the short form, we also call it BFS traversal.

A binary tree is organized into different levels where the root node is at the topmost level (0th level). So the idea of level order traversal is: We start by processing the root node, then process all nodes at the first level, second level, and so on. In other words, we explore all nodes at the current level before moving on to nodes at the next level.

Recursive approach of BFS or level order traversal

This is a brute force idea, where we move from the top to the bottommost level using a loop and process nodes at each level using recursion. The idea looks simple, but implementation would be a little tricky.

The objective to discuss this approach is related to problem-solving. In the recursive solution of a few tree problems, sometimes we pass extra parameters or use helper functions to generate the correct output.

Let’s think about the implementation of recursive BFS traversal:

• The number of levels is equal to the height of the tree. So we first calculate the height of the tree (h) using the function height(root).
• Now we run a loop from l = 0 to h - 1 and access each level in the tree. Inside the loop, we use the function processCurrentLevel(root, l) to visit and process nodes at the current level l.

How do we implement the function processCurrentLevel(root, l)? Let's think! In a standard level-order traversal, we traverse the nodes from left to right at any given level. So we can split the problem into two parts:

1) Traversing all nodes at distance l in the left subtree: We first recursively process all nodes at level l in the left subtree. For this, we call the same function with root->left and l - 1 as function parameters. Why? Because if the current level is l distance from the root, then it would be l - 1 distance away from root->left.

2) Traversing all the nodes at distance l in the right subtree: Now we recursively process all nodes at level l in the right subtree. For this, we call the same function with root->right and l - 1 as function parameters. Why? Because if the current level is l distance from the root, then it would be l - 1 distance away from root->right.

At each step of the recursive call, we are moving one level down. When l == 0, we will be at a node that is distance l from the root. So we process this node. In this way, we can access all nodes at level l recursively.

Implementation code C++

//Calculate the height of the tree
int height(TreeNode* root) 
{
    if (root == NULL)
        return 0;
    else 
    {
        int leftHeight = height(root->left);
        int rightHeight = height(root->right);

        if (leftHeight > rightHeight)
            return (leftHeight + 1);
        else
            return (rightHeight + 1);
    }
}

//Process nodes at the level l
void processCurrentLevel(TreeNode* root, int l) 
{
    if (root == NULL)
        return;
  
    if (l == 0) 
        cout<< root->data;
    else if (l > 0) 
    {
        processCurrentLevel(root->left, l - 1);
        processCurrentLevel(root->right, l - 1);
    }
}

//Traverse nodes in level order traversal
void recursiveLevelOrder(TreeNode* root) 
{
    int h = height(root);
    for (int l = 0; l < h; l = l + 1)
        processCurrentLevel(root, l);
}

Implementation code Python

def height(root):
    if root is None:
        return 0
    else:
        left_height = height(root.left)
        right_height = height(root.right)

        if left_height > right_height:
            return left_height + 1
        else:
            return right_height + 1

def process_current_level(root, l):
    if root is None:
        return
    if l == 0:
        print(root.data)
    elif l > 0:
        process_current_level(root.left, l - 1)
        process_current_level(root.right, l - 1)

def recursive_level_order(root):
    h = height(root)
    for l in range(h):
        process_current_level(root, l)

Implementation code Java

public class LevelOrderTraversal 
{
    // Calculate the height of the tree
    private int height(TreeNode root) 
    {
        if (root == null)
            return 0;
        else 
        {
            int leftHeight = height(root.left);
            int rightHeight = height(root.right);
    
            if (leftHeight > rightHeight)
                return (leftHeight + 1);
            else
                return (rightHeight + 1);
        }
    }
    
    // Process nodes at the level l
    private void processCurrentLevel(TreeNode root, int l) 
    {
        if (root == null)
            return;
      
        if (l == 0)
            System.out.print(root.data + " ");
        else if (l > 0) 
        {
            processCurrentLevel(root.left, l - 1);
            processCurrentLevel(root.right, l - 1);
        }
    }
    
    // Traverse nodes in level order traversal
    public void recursiveLevelOrder(TreeNode root) 
    {
        int h = height(root);
        for (int l = 0; l < h; l = l + 1)
            processCurrentLevel(root, l);
    }
}

Time and space complexity analysis

To process nodes at a specific level l, we recursively call the function processCurrentLevel() for both the left and right subtree. In other words, for each level l, we traverse l - 1 level down and recursively access each node from level 0 to level l. So one idea becomes clear: the total number of operations depends on the height of the tree.

The worst-case scenario would be a skewed tree where each node has only one child, resulting in a tree height of O(n). In this case, processCurrentLevel() takes O(n) time for the last level, O(n-1) time for the second-last level, and so on. So time complexity = O(n) + O(n - 1) + .. + O(1) = O(n + n-1 + ... + 1) = O(n^2). What about the best-case scenario? Think and explore!

The space complexity depends on the size of the recursion call stack, which is equal to the height of the tree. So skewed tree will be the scenario of the worst case and recursion will use O(n) space for the call stack. The space complexity in the worst case = O(n).

Efficient approach: BFS or level order traversal using queue

Now critical questions are: Can we optimize the time complexity of BFS traversal? Can we traverse the tree level by level in O(n) time complexity? Let's think! If we observe the order of nodes in level order traversal:

• We first process the root node at level 0, and then we process the left and right child at level 1 (assuming left to right order).
• Similarly, at the second level, we first process children of the left child of the root then process children of the right child. This process goes on for all levels.

So one idea is clear: If we first process a node x at any given level, then the children of node x will be processed before the children of any other node at the next level. In other words, if node x comes before node y at the same level, then at the next level, we will process the children of node x before processing the children of node y. This is the First In First Out (FIFO) order of processing nodes. So we can use a queue data structure to simulate the level order or breadth-first search (BFS) traversal.

Implementation steps: BFS traversal using queue

Step 1: We create an empty queue treeQueue and initialize it with the root node.

Step 2: Now we run a loop until the treeQueue is empty. Inside the loop, we declare a variable called currNode to keep track of the current node during the traversal.

• We start the loop by removing the front node from the treeQueue and assigning it to currNode. 
• Now we process currNode and insert its left and right child nodes into the treeQueue if they exist: 1) If the left child of currNode is not NULL, we insert it into the treeQueue 2) If the right child of currNode is not NULL, we insert it into the treeQueue.

Step 3: After processing the rightmost node at the last level, there are no more nodes inside the queue to process. We exit the loop, and the level order traversal is complete.

Implementation code  C++

void iterativeLevelOrder(TreeNode* root) 
{
    if (root == NULL) 
        return;

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

    while (treeQueue.empty() == false) 
    {
        TreeNode* currNode = treeQueue.front();
        treeQueue.pop();
        cout << currNode->data << " ";
      
        if (currNode->left != NULL) 
            treeQueue.push(currNode->left);
            
        if (currNode->right != NULL) 
            treeQueue.push(currNode->right);
    }
}

Implementation code  Python

def iterativeLevelOrder(root):
    if root is None:
        return
    
    treeQueue = []
    treeQueue.append(root)

    while len(treeQueue) > 0:
        currNode = treeQueue.pop(0)
        print(currNode.data, end=" ")

        if currNode.left is not None:
            treeQueue.append(currNode.left)

        if currNode.right is not None:
            treeQueue.append(currNode.right)

Implementation code  Java

class BFSTreeTraversal 
{
    public void iterativeLevelOrder(TreeNode root) 
    {
        if (root == null)
            return;

        Queue<TreeNode> treeQueue = new LinkedList<>();
        treeQueue.add(root);

        while (treeQueue.isEmpty() == false) 
        {
            TreeNode currNode = treeQueue.poll();
            System.out.print(currNode.data + " ");

            if (currNode.left != null)
                treeQueue.add(currNode.left);
            
            if (currNode.right != null)
                treeQueue.add(currNode.right);
        }
    }
}

Time complexity analysis of the BFS traversal

Suppose n number of nodes are given in the input. 

• The time complexity of each enqueue and dequeue operation = O(1).
• We are doing two queue operations for each node inside the loop: Inserting once into the queue and deleting once from the queue. So total queue operations = 2n.
• Overall time complexity = Total queue operations * Time complexity of each queue operation = 2n * O(1) = O(n)

Space complexity analysis of the BFS traversal

Space complexity is equal to the queue size. We process nodes level by level, so the max queue size depends on the level with the maximum number of nodes or max-width of the binary tree. If the maximum width of the binary tree is w, then space complexity = O(w). The idea is simple: w depends on the structure of a given binary tree. How? Let’s think!

Worst case: When tree is balanced

When tree is balanced, the last level will have maximum width or maximum number of nodes, which will be 2^h (where h is the height of the tree). For balanced tree, h = logn and required size of the queue = O(2^h) = O(2^(log n)) = O(n). Space complexity = O(n).

Best case: When tree is skewed

In such case, every level will have maximum of one node. So at any point, there will be at most one node in the queue. So required size of the queue = O(1). Space complexity = O(1).

BFS vs DFS traversal of binary tree

• Both traversals require O(n) time as they visit every node exactly once.
• Depth-first traversal starts from the root, goes to the depth as far as possible, and then backtracks from there. In other words, it visits nodes from bottom of the tree. But in breadth-first search, we explore nodes level by level i.e. in order of their distance from the root node. So if our problem is to search for something closer to the root, we would prefer BFS. And if we need to search for something in the depth of tree or node closer to leaf, we would prefer DFS.
• In BFS traversal, we use queue data structure to store nodes of different levels. But in DFS traversal, we use the stack (If recursive, system use call stack) to store all ancestors of a node.
• The memory taken by both BFS and DFS depends on the structure of tree. Extra space required for BFS Traversal is O(w), but additional space needed for DFS Traversal is O(h). Here w = maximum width of binary tree and h = maximum height of binary tree. In the worst case, both require O(n) extra space, but worst cases occur for different types of trees. For example, space needed for BFS Traversal is likely to be more when a tree is more balanced, and extra space for DFS Traversal is likely to be more when a tree is less balanced.
• Sometimes, when node order is not required in problem-solving, we can use both BFS and DFS traversals. But in some cases, such things are not possible. We need to identify the use case of traversal to solve the problems efficiently.

Problems to practice using BFS traversal

• Level order traversal in spiral form
• Reverse Level Order Traversal
• Left View of Binary Tree
• Maximum width of a binary tree
• Min Depth of Binary Tree
• Level with the maximum number of nodes
• Count the number of nodes at a given level
• Convert a binary tree into a mirror tree

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