# 979. Distribute Coins in Binary Tree

## Description

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.


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.

## Solutions

• /**
* 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;
*     }
* }
*/
class Solution {
private int ans;

public int distributeCoins(TreeNode root) {
dfs(root);
return ans;
}

private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int left = dfs(root.left);
int right = dfs(root.right);
ans += Math.abs(left) + Math.abs(right);
return left + right + root.val - 1;
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int distributeCoins(TreeNode* root) {
int ans = 0;
function<int(TreeNode*)> dfs = [&](TreeNode* root) -> int {
if (!root) {
return 0;
}
int left = dfs(root->left);
int right = dfs(root->right);
ans += abs(left) + abs(right);
return left + right + root->val - 1;
};
dfs(root);
return ans;
}
};

• # 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:
def dfs(root):
if root is None:
return 0
left, right = dfs(root.left), dfs(root.right)
nonlocal ans
ans += abs(left) + abs(right)
return left + right + root.val - 1

ans = 0
dfs(root)
return ans


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func distributeCoins(root *TreeNode) (ans int) {
var dfs func(*TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
left, right := dfs(root.Left), dfs(root.Right)
ans += abs(left) + abs(right)
return left + right + root.Val - 1
}
dfs(root)
return
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function distributeCoins(root: TreeNode | null): number {
let ans = 0;
const dfs = (root: TreeNode | null) => {
if (!root) {
return 0;
}
const left = dfs(root.left);
const right = dfs(root.right);
ans += Math.abs(left) + Math.abs(right);
return left + right + root.val - 1;
};
dfs(root);
return ans;
}