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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0501_0600.s0543_diameter_of_binary_tree;

// #Easy #Top_100_Liked_Questions #Depth_First_Search #Tree #Binary_Tree #Level_2_Day_7_Tree
// #Udemy_Tree_Stack_Queue #Big_O_Time_O(n)_Space_O(n)
// #2022_08_02_Time_1_ms_(65.86%)_Space_43.5_MB_(33.52%)

import com_github_leetcode.TreeNode;

/*
 * Definition for a binary tree node.
 * 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;
 *     }
 * }
 */
/**
 * 543 - Diameter of Binary Tree\.
 *
 * Easy
 *
 * Given the `root` of a binary tree, return _the length of the **diameter** of the tree_.
 *
 * The **diameter** of a binary tree is the **length** of the longest path between any two nodes in a tree. This path may or may not pass through the `root`.
 *
 * The **length** of a path between two nodes is represented by the number of edges between them.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/03/06/diamtree.jpg)
 *
 * **Input:** root = [1,2,3,4,5]
 *
 * **Output:** 3
 *
 * **Explanation:** 3 is the length of the path [4,2,1,3] or [5,2,1,3]. 
 *
 * **Example 2:**
 *
 * **Input:** root = [1,2]
 *
 * **Output:** 1 
 *
 * **Constraints:**
 *
 * *   The number of nodes in the tree is in the range [1, 10⁴].
 * *   `-100 <= Node.val <= 100`
**/
public class Solution {
    private int diameter;

    public int diameterOfBinaryTree(TreeNode root) {
        diameter = 0;
        diameterOfBinaryTreeUtil(root);
        return diameter;
    }

    private int diameterOfBinaryTreeUtil(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftLength = root.left != null ? 1 + diameterOfBinaryTreeUtil(root.left) : 0;
        int rightLength = root.right != null ? 1 + diameterOfBinaryTreeUtil(root.right) : 0;
        diameter = Math.max(diameter, leftLength + rightLength);
        return Math.max(leftLength, rightLength);
    }
}