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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0501_0600.s0572_subtree_of_another_tree;

// #Easy #Depth_First_Search #Tree #Binary_Tree #Hash_Function #String_Matching
// #Algorithm_II_Day_7_Breadth_First_Search_Depth_First_Search
// #2022_08_10_Time_1_ms_(100.00%)_Space_47.2_MB_(13.44%)

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;
 *     }
 * }
 */
/**
 * 572 - Subtree of Another Tree\.
 *
 * Easy
 *
 * Given the roots of two binary trees `root` and `subRoot`, return `true` if there is a subtree of `root` with the same structure and node values of `subRoot` and `false` otherwise.
 *
 * A subtree of a binary tree `tree` is a tree that consists of a node in `tree` and all of this node's descendants. The tree `tree` could also be considered as a subtree of itself.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/04/28/subtree1-tree.jpg)
 *
 * **Input:** root = [3,4,5,1,2], subRoot = [4,1,2]
 *
 * **Output:** true
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/04/28/subtree2-tree.jpg)
 *
 * **Input:** root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]
 *
 * **Output:** false
 *
 * **Constraints:**
 *
 * *   The number of nodes in the `root` tree is in the range `[1, 2000]`.
 * *   The number of nodes in the `subRoot` tree is in the range `[1, 1000]`.
 * *   -10⁴ <= root.val <= 10⁴
 * *   -10⁴ <= subRoot.val <= 10⁴
**/
public class Solution {
    private boolean isSubtreeFound(TreeNode root, TreeNode subRoot) {
        if (root == null && subRoot == null) {
            return true;
        }
        if (root == null || subRoot == null) {
            return false;
        }
        if (root.val == subRoot.val) {
            return isSubtreeFound(root.left, subRoot.left) && isSubtree(root.right, subRoot.right);
        } else {
            return false;
        }
    }

    public boolean isSubtree(TreeNode root, TreeNode subRoot) {
        if (root == null && subRoot == null) {
            return true;
        }
        if (root == null || subRoot == null) {
            return false;
        }
        return isSubtreeFound(root, subRoot)
                || isSubtree(root.left, subRoot)
                || isSubtree(root.right, subRoot);
    }
}