# 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`.

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def distributeCoins(self, root: Optional[TreeNode]) -> int:
        self.result = 0
        self.dfs(root)
        return self.result
    
    def dfs(self, root):
        if root is None:
            return 0
        left, right = self.dfs(root.left) , self.dfs(root.right)
        self.result += abs(left) + abs(right)
        return root.val + left + right - 1
```