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;
    }
}