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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0901_1000.s0979_distribute_coins_in_binary_tree;

// #Medium #Depth_First_Search #Tree #Binary_Tree
// #2022_03_31_Time_0_ms_(100.00%)_Space_41.8_MB_(85.81%)

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;
 *     }
 * }
 */
/**
 * 979 - Distribute Coins in Binary Tree\.
 *
 * Medium
 *
 * You are given the `root` of a binary tree with `n` nodes where each `node` in the tree has `node.val` coins. There are `n` coins in total throughout the whole tree.
 *
 * In one move, we may choose two adjacent nodes and move one coin from one node to another. A move may be from parent to child, or from child to parent.
 *
 * Return _the **minimum** number of moves required to make every node have **exactly** one coin_.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2019/01/18/tree1.png)
 *
 * **Input:** root = [3,0,0]
 *
 * **Output:** 2
 *
 * **Explanation:** From the root of the tree, we move one coin to its left child, and one coin to its right child.
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2019/01/18/tree2.png)
 *
 * **Input:** root = [0,3,0]
 *
 * **Output:** 3
 *
 * **Explanation:** From the left child of the root, we move two coins to the root [taking two moves]. Then, we move one coin from the root of the tree to the right child.
 *
 * **Constraints:**
 *
 * *   The number of nodes in the tree is `n`.
 * *   `1 <= n <= 100`
 * *   `0 <= Node.val <= n`
 * *   The sum of all `Node.val` is `n`.
**/
public class Solution {
    private int num = 0;

    public int distributeCoins(TreeNode root) {
        helper(root);
        return num;
    }

    private int helper(TreeNode node) {
        if (node == null) {
            return 0;
        }
        int total = node.val + helper(node.left) + helper(node.right);
        int leftover = total - 1;
        num += Math.abs(leftover);
        return leftover;
    }
}