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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0601_0700.s0687_longest_univalue_path;

// #Medium #Depth_First_Search #Tree #Binary_Tree
// #2022_03_22_Time_3_ms_(79.43%)_Space_68.5_MB_(25.41%)

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;
 *     }
 * }
 */
/**
 * 687 - Longest Univalue Path\.
 *
 * Medium
 *
 * Given the `root` of a binary tree, return _the length of the longest path, where each node in the path has the same value_. This path may or may not pass through the root.
 *
 * **The length of the path** between two nodes is represented by the number of edges between them.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/10/13/ex1.jpg)
 *
 * **Input:** root = [5,4,5,1,1,5]
 *
 * **Output:** 2
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/10/13/ex2.jpg)
 *
 * **Input:** root = [1,4,5,4,4,5]
 *
 * **Output:** 2
 *
 * **Constraints:**
 *
 * *   The number of nodes in the tree is in the range [0, 10⁴].
 * *   `-1000 <= Node.val <= 1000`
 * *   The depth of the tree will not exceed `1000`.
**/
public class Solution {
    public int longestUnivaluePath(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int[] res = new int[1];
        preorderLongestSinglePathLen(root, res);
        return res[0];
    }

    private int preorderLongestSinglePathLen(TreeNode root, int[] res) {
        if (root == null) {
            return -1;
        }
        int left = preorderLongestSinglePathLen(root.left, res);
        int right = preorderLongestSinglePathLen(root.right, res);

        if (root.left == null || root.val == root.left.val) {
            left = left + 1;
        } else {
            left = 0;
        }

        if (root.right == null || root.val == root.right.val) {
            right = right + 1;
        } else {
            right = 0;
        }
        int longestPathLenPassingThroughRoot = left + right;
        res[0] = Math.max(res[0], longestPathLenPassingThroughRoot);
        return Math.max(left, right);
    }
}