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;
    }
}

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