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MD ARIFUL HAQUE

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# 979. Distribute Coins in Binary Tree

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:

• 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: oot = [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`.

Solution:

``````/**
* Definition for a binary tree node.
* class TreeNode {
*     public \$val = null;
*     public \$left = null;
*     public \$right = null;
*     function __construct(\$val = 0, \$left = null, \$right = null) {
*         \$this->val = \$val;
*         \$this->left = \$left;
*         \$this->right = \$right;
*     }
* }
*/
class Solution {

/**
* @param TreeNode \$root
* @return Integer
*/
function distributeCoins(\$root) {
\$totalMoves = 0;
\$dfs = function(\$node) use (&\$dfs, &\$totalMoves) {
if (!\$node) {
return 0;
}
\$leftMoves = \$dfs(\$node->left);
\$rightMoves = \$dfs(\$node->right);
\$totalMoves += abs(\$leftMoves) + abs(\$rightMoves);
return \$leftMoves + \$rightMoves + \$node->val - 1;
};
\$dfs(\$root);
return \$totalMoves;
}
}
``````