Easy
Given the root
of a binary tree, return its maximum depth.
A binary tree’s maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: 3
Example 2:
Input: root = [1,null,2]
Output: 2
Example 3:
Input: root = []
Output: 0
Example 4:
Input: root = [0]
Output: 1
Constraints:
[0, 104]
.-100 <= Node.val <= 100
To solve the “Maximum Depth of Binary Tree” problem in Java with a Solution
class, we’ll perform a depth-first search (DFS) traversal of the binary tree. Below are the steps:
Create a Solution
class: Define a class named Solution
to encapsulate our solution methods.
Create a maxDepth
method: This method takes the root node of the binary tree as input and returns its maximum depth.
Check for null root: Check if the root is null. If it is, return 0 as the depth.
Perform DFS traversal: Recursively compute the depth of the left and right subtrees. The maximum depth of the binary tree is the maximum depth of its left and right subtrees, plus 1 for the current node.
Return the result: After the DFS traversal is complete, return the maximum depth of the binary tree.
Here’s the Java implementation:
class Solution {
public int maxDepth(TreeNode root) {
if (root == null) return 0; // Check for empty tree
int leftDepth = maxDepth(root.left); // Compute depth of left subtree
int rightDepth = maxDepth(root.right); // Compute depth of right subtree
return Math.max(leftDepth, rightDepth) + 1; // Return maximum depth of left and right subtrees, plus 1 for the current node
}
// Definition for a TreeNode
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {}
TreeNode(int val) { this.val = val; }
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
}
This implementation follows the steps outlined above and efficiently computes the maximum depth of the binary tree in Java using DFS traversal.