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Introduction to Leaf Processing Pattern
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Leaf Processing Pattern focuses on operations involving the leaves of a binary tree. In this pattern, we analyze and manipulate the tree's leaf nodes, which have no children. This approach is useful in various problems where we need to extract information specifically from the leaves, such as finding the sum of all left leaves or counting the total number of leaves.

Here, the overall approach involves traversing the binary tree using techniques like DFS (Depth-First Search) or BFS (Breadth-First Search), and identifying the leaf nodes and performing the required operations. The required operations on the leaf node can be summing their values, etc. Efficient use of recursion or iterative methods helps in processing the leaves while maintaining optimal time and space complexity.

Finding the Leaves of Binary Tree

Now, we will learn to find leaf nodes in a binary tree using two different approaches: Depth-First Search (DFS) and Breadth-First Search (BFS). These methods help in identifying all the leaf nodes, which are the nodes with no children. Let's consider a few sample examples to understand this better.

Example 1:

Given a binary tree:

Image

In this tree, the leaf nodes are 4, 5, and 3, as they do not have any children.

Example 2:

Consider another binary tree:

Image

Here, the leaf nodes are 4 and 5. We will use DFS and BFS to find these leaf nodes in our solutions.

Using the DFS Approach

To use DFS approach, we start at the root node and traverse the tree by recursively visiting each child node. Whenever a node is found with no left or right child, it is identified as a leaf node and added to our result list. The recursion ensures that each path from the root to the leaf is explored completely before backtracking, making it a straightforward and effective way to identify all leaf nodes in the tree. This approach is efficient due to its simplicity and because it leverages the natural structure of the tree, reducing the need for additional data structures.

Step-by-Step Algorithm

  • Step 1: Start at the root node of the binary tree.
  • Step 2: If the current node is null, return (base case for recursion).
  • Step 3: Check if the current node is a leaf node (both left and right children are null).
    • a. If yes, add it to the list of leaf nodes.
  • Step 4: Recursively call the function for the left child of the current node.
  • Step 5: Recursively call the function for the right child of the current node.
  • Step 6: Continue until all nodes have been visited.

Algorithm Walkthrough

Let's walk through the DFS approach step-by-step using Example 1:

Image

  • Initial State:

    • Current Node: 1
    • Leaf Nodes List: []

  • Step 1: Start at the root node (1). It's not a leaf, so we move to its children.

    • Current Node: 1
    • Leaf Nodes List: []

  • Step 2: Recurse to the left child (2). It's not a leaf, so we continue exploring its children.

    • Current Node: 2
    • Leaf Nodes List: []

  • Step 3: Recurse to the left child of 2 (4). This is a leaf node (no children), so we add 4 to the leaf list.

    • Current Node: 4 (Leaf Node)
    • Leaf Nodes List: [4]

  • Step 4: Return to node 2 and recurse to its right child (5). This is also a leaf node, so we add 5 to the leaf list.

    • Current Node: 5 (Leaf Node)
    • Leaf Nodes List: [4, 5]

  • Step 5: Return to the root node (1) and recurse to its right child (3). This is a leaf node, so we add 3 to the leaf list.

    • Current Node: 3 (Leaf Node)
    • Leaf Nodes List: [4, 5, 3]

  • End State:

    • All nodes have been visited.
    • Final Leaf Nodes List: [4, 5, 3].

Code

Python3
Python3
. . . .

Complexity Analysis

  • Time Complexity: The time complexity of this algorithm is O(N), where N is the number of nodes in the binary tree. This is because each node is visited exactly once during the DFS traversal.
  • Space Complexity: The space complexity is O(H), where H is the height of the binary tree. This is due to the maximum depth of the recursion stack. In the worst case, where the tree is skewed, this could be up to O(N).

Using the BFS Approach

Breadth-First Search (BFS), similar to level-order traversal, processes the binary tree level by level, starting from the root and moving to each level’s children.

To use the BFS approach, we start by using a queue to keep track of nodes at each level. We begin from the root and add each node's children to the queue. For every node, we check whether it is a leaf (i.e., it has no children). If it is, we add it to the result list. This approach is effective because it systematically processes each level and ensures all leaf nodes are identified without redundant checks. It is particularly useful when the tree is wide or balanced, as BFS efficiently handles all nodes at a given level before moving deeper.

Step-by-Step Algorithm

  • Step 1: Initialize an empty queue and add the root node to it.
  • Step 2: Initialize an empty list to store leaf nodes.
  • Step 3: While the queue is not empty:
    • a. Remove (dequeue) the front node from the queue.
    • b. Check if this node is a leaf node (both left and right children are null).
      • i. If yes, add it to the list of leaf nodes.
    • c. If the node has a left child, add it to the queue.
    • d. If the node has a right child, add it to the queue.
  • Step 4: Continue until the queue is empty.
  • Step 5: Return the list of leaf nodes.

Algorithm Walkthrough

Let's walk through the BFS approach using Example 1:

Image

  • Initial state:

    • Queue: [1]
    • Leaf Nodes List: []

  • Step 1: Dequeue node 1 from the queue. Queue becomes [].

    • Node 1 is not a leaf.
    • Enqueue its left child 2. Queue becomes [2].
    • Enqueue its right child 3. Queue becomes [2, 3].
    • Leaf Nodes List: []

  • Step 2: Dequeue node 2 from the queue. Queue becomes [3].

    • Node 2 is not a leaf.
    • Enqueue its left child 4. Queue becomes [3, 4].
    • Enqueue its right child 5. Queue becomes [3, 4, 5].
    • Leaf Nodes List: []

  • Step 3: Dequeue node 3 from the queue. Queue becomes [4, 5].

    • Node 3 is a leaf. Add 3 to the leaf list.
    • Leaf Nodes List: [3]

  • Step 4: Dequeue node 4 from the queue. Queue becomes [5].

    • Node 4 is a leaf. Add 4 to the leaf list.
    • Leaf Nodes List: [3, 4]

  • Step 5: Dequeue node 5 from the queue. Queue becomes [].

    • Node 5 is a leaf. Add 5 to the leaf list.
    • Leaf Nodes List: [3, 4, 5]

  • End state:

    • Queue is empty.
    • Final Leaf Nodes List: [3, 4, 5].

Code

Python3
Python3
. . . .

Complexity Analysis

  • Time Complexity: The time complexity of this algorithm is O(N), where N is the number of nodes in the binary tree. This is because each node is visited exactly once during the traversal.

  • Space Complexity: The space complexity is O(W), where W is the maximum width of the binary tree. The space is primarily used by the queue, which, in the worst case, may need to store all nodes at the widest level of the tree. In a balanced binary tree, this is proportional to O(N) in the worst case.

Now, let's start solving the Binary tree leaf related problems.

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