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.

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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.


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Constraints:


	The number of nodes in the tree is n.
	1 &lt;= n &lt;= 100
	0 &lt;= Node.val &lt;= n
	The sum of all Node.val is n.


# 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 not root:
                return 0,0
            
            net_left,left_walk = dfs(root.left)
            net_right,right_walk = dfs(root.right)
            
			# if any node has extra or deficiency in both cases there has to be a walk of abs(extra) or abs(deficiency)
			
            return net_left+net_right+(root.val-1), left_walk+right_walk+abs(net_left)+abs(net_right)
        return dfs(root)[1]
class Solution {
    int count = 0;
        public int helper(TreeNode root)
        {
            if(root == null)
                return 0;
            int left = helper(root.left);
            int right = helper(root.right);
            count+= Math.abs(left)+Math.abs(right);
            return (left+right+root.val-1);
        }
        public int distributeCoins(TreeNode root) {
            helper(root);
            return count;
        }
}
class Solution {
public:
    int ans = 0;
    int recr(TreeNode *root)
    {
        if(!root) return 0;
        int left = recr(root->left);
        int right = recr(root->right);
        int curr = root->val + left + right;
        ans += abs(curr-1);
        return curr-1;
    }
    int distributeCoins(TreeNode* root) {
        recr(root);
        return ans;
    }
};
var distributeCoins = function(root) {
    var moves = 0;
    postorder(root);
    return moves;

    function postorder(node){
        if(!node)
            return 0;

        const subTotal = postorder(node.left) + postorder(node.right);
        const result = node.val - 1 + subTotal;
        moves += Math.abs(result);

        return result;

    }
};
Distribute Coins in Binary Tree