Introduction to Tree Depth Pattern

The Tree Depth Pattern is crucial in binary tree problems where understanding the depth of the tree is key. Depth refers to the number of nodes along the path from the root to the deepest or shallowest leaf. Problems involving depth often require finding the minimum or maximum depth of the tree, which can be important for understanding the tree’s structure, balance, and overall health. Mastering this pattern will help you solve various problems efficiently by focusing on how deep or shallow different parts of the tree are.

To approach these problems, the general strategy involves traversing the tree while keeping track of the depth at each level. There are two common methods:

• Breadth-First Search (BFS): This method processes nodes level by level, making it easier to calculate the depth of each level. It's particularly useful for finding the minimum depth of the tree.

• Depth-First Search (DFS): This method explores as far as possible along each branch before backtracking. It's often used for calculating the maximum depth and is ideal for a more recursive approach.

By understanding and applying these techniques, you'll be able to handle problems related to tree depth effectively, ensuring that you can identify the key characteristics of the tree structure quickly.

Now, let's understand to find the depth of particular node via example below.

Problem Statement

Given a root of the binary tree and node value n, return the depth of the node having value n in the tree. If the node does not exist in the tree, return -1.

Examples

Example 1:

• Input: root = [3, 9, 20, null, null, 15, 7], n = 15

• Expected Output: 3
• Explanation: The node with value 15 is at depth 2.

Example 2:

• Input: root = [1, 2, 3, 4, 5, null, null, 8], n = 8

• Expected Output: 4
• Explanation: The node with value 8 is at depth 4.

Finding a Tree Depth Using the DFS

Here, the goal is to find the depth of a specific node by exploring the tree's structure, starting from the root and moving down each branch. DFS is effective here because it allows us to explore each possible path to the node, ensuring we find the exact depth. By keeping track of the current depth at each level, we can determine the node's position in the tree. This approach is efficient as it only traverses paths that are necessary to find the target node, making it optimal for solving this problem.

Step-by-Step Algorithm

Here’s a detailed explanation of the DFS algorithm to find the depth of a given node in a binary tree:

• Step 1: Start at the root of the binary tree with a depth initialized to 1. This is because the root node is at depth 1 (as per the provided code).

• Step 2: Implement a recursive DFS function that takes three parameters: the current node (TreeNode), the target node value (int), and the current depth (int).

• Step 3: Base Case: Check if the current node is null. If it is, return -1 because the target node is not found in this path.

• Step 4: Check Target: If the current node’s value matches the target value, return the current depth.

• Step 5: Recurse Left: Call the DFS function recursively on the left child of the current node, increasing the depth by 1.

• Step 6: Check Left Subtree: If the target node is found in the left subtree (i.e., the returned depth is not -1), return the depth from the left subtree.

• Step 7: Recurse Right: If the node is not found in the left subtree, call the DFS function recursively on the right child of the current node, also increasing the depth by 1.

• Step 8: Return Depth: If the target node is found in the right subtree, return the depth. If the node is not found in either subtree, return -1.

Algorithm Walkthrough

Let's walk through the algorithm with a real input to see how it works:

Walkthrough:

• Step 1: Start at the root node (3) with a depth of 1.

• Step 2: The current node (3) does not match the target value (15).

• Step 3: Recur on the left child (9) with depth 2.

  • The node 9 does not match 15, and it has no children, so return -1.

• Step 4: Recur on the right child (20) with depth 2.

  • The node 20 does not match 15. Recur on its left child (15) with depth 3.

  • The node 15 matches the target value, so return 3.

• Step 5: The algorithm returns 3 as the depth of the node with value 15.

Code

// class TreeNode {
//     int val;
//     TreeNode left, right;

//     TreeNode(int val) {
//         this.val = val;
//         this.left = null;
//         this.right = null;
//     }
// }

public class Solution {


    public int findDepth(TreeNode root, int target) {
        return dfs(root, target, 1); // Start DFS from root at depth 1
    }

    private int dfs(TreeNode node, int target, int depth) {
        // Base case: if node is null, return -1 (target not found)
        if (node == null) {
            return -1;
        }

        // If current node is the target, return current depth
        if (node.val == target) {
            return depth;
        }

        // Recur for the left subtree
        int leftDepth = dfs(node.left, target, depth + 1);
        if (leftDepth != -1) {
            return leftDepth; // Target found in left subtree
        }

        // Recur for the right subtree if not found in left
        return dfs(node.right, target, depth + 1); // Check right subtree
    }

    public static void main(String[] args) {
        // Creating the binary tree
        TreeNode root = new TreeNode(3);
        root.left = new TreeNode(9);
        root.right = new TreeNode(20);
        root.right.left = new TreeNode(15);
        root.right.right = new TreeNode(7);

        // Creating an instance of the Solution class
        Solution solution = new Solution();

        // Test case 1: Finding depth of node with value 15
        int depth1 = solution.findDepth(root, 15);
        System.out.println("Depth of node 15: " + depth1); // Expected output: 3
    }
}

Complexity Analysis

• Time Complexity: The time complexity of the DFS approach in this problem is , where N is the number of nodes in the binary tree. This is because, in the worst case, we may need to explore all nodes to find the target node, especially if the node is located deep in the tree.

• Space Complexity: The space complexity is , where H is the height of the binary tree. This space is required for the call stack due to the recursive nature of DFS. In the worst case, the height of the tree could be equal to the number of nodes (in a skewed tree), making the space complexity .

Finding a Tree Depth Using the BFS

The BFS also is ideal for this type of problem because it explores the tree level by level, making it well-suited for finding the shortest path or depth in a binary tree. By processing all nodes at one depth level before moving to the next, BFS ensures that we find the target node at the shallowest possible depth, if it exists. This approach is effective because it avoids unnecessary deep exploration and directly focuses on the level where the target node resides. Additionally, BFS naturally handles the traversal in a way that keeps track of the depth, making it easy to return the correct depth as soon as the target node is found.

Step-by-Step Algorithm

1. Check for an empty tree:

   • If root is null, return -1.

2. Initialize the queue:

   • Create a queue and add the root node with depth 1.

3. Begin BFS traversal:

   • While the queue is not empty, do the following:
     • Determine the number of nodes at the current level (levelSize).

4. Process each node at the current level:

   • For each node in the current level:
     • Dequeue the node.
     • If the node’s value matches the target, return the current depth.
     • Add the left and right children (if they exist) to the queue with an incremented depth.

5. Increment depth:

   • After processing all nodes at the current level, increment the depth by 1.

6. Return -1:

   • If the queue is empty and the target node was not found, return -1.

Algorithm Walkthrough

Initial State:

• Queue: [3]
• Depth: 1

Steps:

1. Level 1:

   • Queue before processing: [3]
   • Dequeue 3.
   • 3 does not match 15.
   • Add 9 and 20 to the queue.
   • Queue after processing: [9, 20]
   • Increment depth to 2.
   • Depth: 2

2. Level 2:

   • Queue before processing: [9, 20]
   • Dequeue 9.
   • 9 does not match 15. It has no children.
   • Queue after processing 9: [20]
   • Dequeue 20.
   • 20 does not match 15.
   • Add 15 and 7 to the queue.
   • Queue after processing 20: [15, 7]
   • Increment depth to 3.
   • Depth: 3

3. Level 3:

   • Queue before processing: [15, 7]
   • Dequeue 15.
   • 15 matches the target node.
   • Return depth: 3

Final Output: The depth of node 15 is 3.

Code

import java.util.LinkedList;
import java.util.Queue;

// class TreeNode {
//     int val;
//     TreeNode left, right;

//     TreeNode(int val) {
//         this.val = val;
//         this.left = null;
//         this.right = null;
//     }
// }


public class Solution {

    // BFS method to find the depth of a given node value
    public int findDepth(TreeNode root, int target) {
        // If the tree is empty, return -1
        if (root == null) {
            return -1;
        }

        // Initialize a queue to store nodes along with their depth
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        int depth = 1; // Start from depth 1

        // Perform BFS traversal
        while (!queue.isEmpty()) {
            int levelSize = queue.size(); // Number of nodes at the current depth level

            // Process all nodes at the current level
            for (int i = 0; i < levelSize; i++) {
                TreeNode currentNode = queue.poll();

                // If the current node matches the target, return the current depth
                if (currentNode.val == target) {
                    return depth;
                }

                // Add left and right children to the queue, if they exist
                if (currentNode.left != null) {
                    queue.add(currentNode.left);
                }
                if (currentNode.right != null) {
                    queue.add(currentNode.right);
                }
            }

            // Increment depth as we move to the next level
            depth++;
        }

        // If the target node is not found, return -1
        return -1;
    }

    // Main method to test the algorithm
    public static void main(String[] args) {
        // Creating the binary tree
        TreeNode root = new TreeNode(3);
        root.left = new TreeNode(9);
        root.right = new TreeNode(20);
        root.right.left = new TreeNode(15);
        root.right.right = new TreeNode(7);

        // Creating an instance of the Solution class
        Solution solution = new Solution();

        // Test case 1: Finding depth of node with value 15
        int depth1 = solution.findDepth(root, 15);
        System.out.println("Depth of node 15: " + depth1); // Expected output: 3
    }
}

Complexity Analysis:

• Time Complexity: The time complexity of the BFS approach in this problem is , where N is the number of nodes in the binary tree. In the worst case, we might need to traverse all nodes to find the target node, particularly if the target is located at the last level of the tree.

• Space Complexity: The space complexity is , where W is the maximum width of the tree. This is because BFS requires storing nodes in a queue, and the maximum space needed occurs when the largest number of nodes are at a single level. In a balanced binary tree, this would be around , but generally, we refer to it as .

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