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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0601_0700.s0606_construct_string_from_binary_tree;

// #Easy #String #Depth_First_Search #Tree #Binary_Tree
// #2022_03_21_Time_1_ms_(100.00%)_Space_42.3_MB_(91.51%)

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;
 *     }
 * }
 */
/**
 * 606 - Construct String from Binary Tree\.
 *
 * Easy
 *
 * Given the `root` of a binary tree, construct a string consisting of parenthesis and integers from a binary tree with the preorder traversal way, and return it.
 *
 * Omit all the empty parenthesis pairs that do not affect the one-to-one mapping relationship between the string and the original binary tree.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/05/03/cons1-tree.jpg)
 *
 * **Input:** root = [1,2,3,4]
 *
 * **Output:** "1(2(4))(3)"
 *
 * **Explanation:** Originally, it needs to be "1(2(4)())(3()())", but you need to omit all the unnecessary empty parenthesis pairs. And it will be "1(2(4))(3)"
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/05/03/cons2-tree.jpg)
 *
 * **Input:** root = [1,2,3,null,4]
 *
 * **Output:** "1(2()(4))(3)"
 *
 * **Explanation:** Almost the same as the first example, except we cannot omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.
 *
 * **Constraints:**
 *
 * *   The number of nodes in the tree is in the range [1, 10⁴].
 * *   `-1000 <= Node.val <= 1000`
**/
public class Solution {
    public String tree2str(TreeNode t) {
        if (t == null) {
            return "";
        }
        StringBuilder sb = new StringBuilder();
        preorder(t, sb);
        return sb.toString();
    }

    private void preorder(TreeNode root, StringBuilder sb) {
        if (root == null) {
            return;
        }
        sb.append(root.val);
        if (root.left != null) {
            sb.append("(");
            preorder(root.left, sb);
            sb.append(")");
        }
        if (root.right != null) {
            if (root.left == null) {
                sb.append("()");
            }
            sb.append("(");
            preorder(root.right, sb);
            sb.append(")");
        }
    }
}