# 26. Distribute Coins in Binary Tree

You are given the `root` of a binary tree with `n` nodes where 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. (A 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:**![](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.
```

**Example 3:**![](https://assets.leetcode.com/uploads/2019/01/18/tree3.png)

```
Input: root = [1,0,2]
Output: 2
```

**Example 4:**![](https://assets.leetcode.com/uploads/2019/01/18/tree4.png)

```
Input: root = [1,0,0,null,3]
Output: 4
```

## Solution: (Postorder traversal)

**Approach:**\
**Keeping track of maximum no of coins moving out from a node and counting moves**<br>

![image](https://assets.leetcode.com/users/votrubac/image_1548011422.png)

```cpp
class Solution
{
public:
    int moves = 0;
    int countCoins(TreeNode *root)
    {
        if (root == NULL)
        {
            return 0;
        }

        int left = countCoins(root->left);
        int right = countCoins(root->right);

        moves += abs(left) + abs(right);
        return root->val + left + right - 1;
    }

    int distributeCoins(TreeNode *root)
    {
        if (root == NULL)
        {
            return 0;
        }

        countCoins(root);
        return moves;
    }
};
```


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