Distribute Coins in Binary Tree

Problem statement

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:

Input: root = [3,0,0]Output: 2Explanation: From the root of the tree, we move one coin to its left child, and one coin to its right child.

Example 2:

Input: root = [0,3,0]Output: 3Explanation: 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.

My solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var distributeCoins = function(root) {
    let moves = 0;
    
    function dfs(node) {
        if (!node) {
            return 0;
        }
        
        const left = dfs(node.left);
        const right = dfs(node.right);
        
        // console.log(node, moves, left, right);
        
        moves += Math.abs(left) + Math.abs(right);
        
        return left + right + node.val - 1;
    }
    
    dfs(root)
    
    return moves;
};