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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0901_1000.s0951_flip_equivalent_binary_trees;

// #Medium #Depth_First_Search #Tree #Binary_Tree
// #2022_12_26_Time_0_ms_(100.00%)_Space_40.2_MB_(73.95%)

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;
 *     }
 * }
 */
/**
 * 951 - Flip Equivalent Binary Trees\.
 *
 * Medium
 *
 * For a binary tree **T** , we can define a **flip operation** as follows: choose any node, and swap the left and right child subtrees.
 *
 * A binary tree **X** is _flip equivalent_ to a binary tree **Y** if and only if we can make **X** equal to **Y** after some number of flip operations.
 *
 * Given the roots of two binary trees `root1` and `root2`, return `true` if the two trees are flip equivalent or `false` otherwise.
 *
 * **Example 1:**
 *
 * ![Flipped Trees Diagram](https://assets.leetcode.com/uploads/2018/11/29/tree_ex.png)
 *
 * **Input:** root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7]
 *
 * **Output:** true
 *
 * **Explanation:** We flipped at nodes with values 1, 3, and 5.
 *
 * **Example 2:**
 *
 * **Input:** root1 = [], root2 = []
 *
 * **Output:** true
 *
 * **Example 3:**
 *
 * **Input:** root1 = [], root2 = [1]
 *
 * **Output:** false
 *
 * **Constraints:**
 *
 * *   The number of nodes in each tree is in the range `[0, 100]`.
 * *   Each tree will have **unique node values** in the range `[0, 99]`.
**/
public class Solution {
    public boolean flipEquiv(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null) {
            return true;
        }
        if (root1 == null || root2 == null) {
            return false;
        }
        if (root1.val != root2.val) {
            return false;
        }
        return flipEquiv(root1.left, root2.left) && flipEquiv(root1.right, root2.right)
                || flipEquiv(root1.left, root2.right) && flipEquiv(root1.right, root2.left);
    }
}