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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0901_1000.s0988_smallest_string_starting_from_leaf;

// #Medium #String #Depth_First_Search #Tree #Binary_Tree
// #2022_03_31_Time_4_ms_(58.47%)_Space_44.1_MB_(60.60%)

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;
 *     }
 * }
 */
/**
 * 988 - Smallest String Starting From Leaf\.
 *
 * Medium
 *
 * You are given the `root` of a binary tree where each node has a value in the range `[0, 25]` representing the letters `'a'` to `'z'`.
 *
 * Return _the **lexicographically smallest** string that starts at a leaf of this tree and ends at the root_.
 *
 * As a reminder, any shorter prefix of a string is **lexicographically smaller**.
 *
 * *   For example, `"ab"` is lexicographically smaller than `"aba"`.
 *
 * A leaf of a node is a node that has no children.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2019/01/30/tree1.png)
 *
 * **Input:** root = [0,1,2,3,4,3,4]
 *
 * **Output:** "dba"
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2019/01/30/tree2.png)
 *
 * **Input:** root = [25,1,3,1,3,0,2]
 *
 * **Output:** "adz"
 *
 * **Example 3:**
 *
 * ![](https://assets.leetcode.com/uploads/2019/02/01/tree3.png)
 *
 * **Input:** root = [2,2,1,null,1,0,null,0]
 *
 * **Output:** "abc"
 *
 * **Constraints:**
 *
 * *   The number of nodes in the tree is in the range `[1, 8500]`.
 * *   `0 <= Node.val <= 25`
**/
public class Solution {
    private String res = "";

    public String smallestFromLeaf(TreeNode root) {
        dfs(root, new StringBuilder());
        return res;
    }

    private void dfs(TreeNode root, StringBuilder currStr) {
        if (root == null) {
            return;
        }
        currStr.insert(0, (char) (root.val + 97));
        if (root.left == null && root.right == null) {
            if (res.equals("")) {
                res = currStr.toString();
            } else {
                res = res.compareTo(currStr.toString()) > 0 ? currStr.toString() : res;
            }
        } else {
            dfs(root.left, currStr);
            dfs(root.right, currStr);
        }
        currStr.deleteCharAt(0);
    }
}