979. Distribute Coins in Binary Tree

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# 題目敘述

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: 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

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.

// 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;

    }

}

# 解題思路

# Solution

class Solution {

    private int moves = 0;

    public int distributeCoins(TreeNode root) {

        moves = 0;

        dfs(root);

        return moves;

    }

    public int dfs(TreeNode root) {

        if(root == null) {

            return 0;

        }

        int left = dfs(root.left);

        int right = dfs(root.right);

        moves += Math.abs(left) + Math.abs(right);

        return root.val + left + right - 1;

    }

}