---
link: https://leetcode.com/problems/distribute-coins-in-binary-tree/
tags: tree, bt, math
---

# 979. Distribute Coins in Binary Tree

## Question
Given the `root` of a binary tree with `N` nodes, each `node` in the tree has `node.val` coins, and there are `N` coins total.

In one move, we may choose two adjacent nodes and move one coin from one node to another. (The move may be from parent to child, or from child to parent.)

Return the number of moves required to make every node have exactly one coin.

**Example 1:**

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Input: [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:**

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Input: [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.

**Example 3:**

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Input: [1,0,2]
Output: 2

**Example 4:**

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Input: [1,0,0,null,3]
Output: 4

**Note:**

1. `1<= N <= 100`
2. `0 <= node.val <= N`

## Solution: Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def distributeCoins(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        self.result = 0
        self.helper(root)
        return self.result
        
    def helper(self, node):
        if node is None:
            return 0
        
        L, R = self.helper(node.left), self.helper(node.right)
        self.result += abs(L) + abs(R)
        return node.val + L + R - 1

We only care about the excess between a child node and its parent. If the child has a value of 0 and has no children of its own, it has an excess of -1 because it requires a coin from its parent. The number of moves is determined my the parent by looking at the excess number of coins from its child. If the excess is -1, then it require 1 move to move a coin from the parent to the child.

Because a node knows the excess of itself and its subtree, we can solve this recursively.

## Solution: Java