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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0872_leaf_similar_trees;

// #Easy #Depth_First_Search #Tree #Binary_Tree
// #2022_03_28_Time_0_ms_(100.00%)_Space_41.8_MB_(64.12%)

import com_github_leetcode.TreeNode;
import java.util.ArrayList;
import java.util.List;
import java.util.Objects;

/*
 * 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;
 *     }
 * }
 */
/**
 * 872 - Leaf-Similar Trees\.
 *
 * Easy
 *
 * Consider all the leaves of a binary tree, from left to right order, the values of those leaves form a **leaf value sequence**_._
 *
 * ![](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/16/tree.png)
 *
 * For example, in the given tree above, the leaf value sequence is `(6, 7, 4, 9, 8)`.
 *
 * Two binary trees are considered _leaf-similar_ if their leaf value sequence is the same.
 *
 * Return `true` if and only if the two given trees with head nodes `root1` and `root2` are leaf-similar.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/09/03/leaf-similar-1.jpg)
 *
 * **Input:** root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null,9,8]
 *
 * **Output:** true
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/09/03/leaf-similar-2.jpg)
 *
 * **Input:** root1 = [1,2,3], root2 = [1,3,2]
 *
 * **Output:** false
 *
 * **Constraints:**
 *
 * *   The number of nodes in each tree will be in the range `[1, 200]`.
 * *   Both of the given trees will have values in the range `[0, 200]`.
**/
public class Solution {
    public boolean leafSimilar(TreeNode root1, TreeNode root2) {
        List list1 = new ArrayList<>();
        List list2 = new ArrayList<>();
        preOrder(root1, list1);
        preOrder(root2, list2);
        // compare the lists
        if (list1.size() != list2.size()) {
            return false;
        }
        for (int i = 0; i < list1.size(); i++) {
            if (!Objects.equals(list1.get(i), list2.get(i))) {
                return false;
            }
        }
        return true;
    }

    private void preOrder(TreeNode root, List list) {
        if (root != null) {
            if (root.left == null && root.right == null) {
                list.add(root.val);
            }
            preOrder(root.left, list);
            preOrder(root.right, list);
        }
    }
}