Add solution #979

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,061 LeetCode solutions in JavaScript
+# 1,062 LeetCode solutions in JavaScript
 
 [https://leetcode.com/](https://leetcode.com/)
 
@@ -787,6 +787,7 @@
 976|[Largest Perimeter Triangle](./solutions/0976-largest-perimeter-triangle.js)|Easy|
 977|[Squares of a Sorted Array](./solutions/0977-squares-of-a-sorted-array.js)|Easy|
 978|[Longest Turbulent Subarray](./solutions/0978-longest-turbulent-subarray.js)|Medium|
+979|[Distribute Coins in Binary Tree](./solutions/0979-distribute-coins-in-binary-tree.js)|Medium|
 985|[Sum of Even Numbers After Queries](./solutions/0985-sum-of-even-numbers-after-queries.js)|Easy|
 989|[Add to Array-Form of Integer](./solutions/0989-add-to-array-form-of-integer.js)|Easy|
 994|[Rotting Oranges](./solutions/0994-rotting-oranges.js)|Medium|

solutions/0979-distribute-coins-in-binary-tree.js

@@ -0,0 +1,43 @@
+/**
+ * 979. Distribute Coins in Binary Tree
+ * https://leetcode.com/problems/distribute-coins-in-binary-tree/
+ * Difficulty: 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.
+ */
+
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {number}
+ */
+var distributeCoins = function(root) {
+  let result = 0;
+  traverse(root);
+  return result;
+
+  function traverse(node) {
+    if (!node) return 0;
+
+    const leftExcess = traverse(node.left);
+    const rightExcess = traverse(node.right);
+    const totalExcess = node.val + leftExcess + rightExcess - 1;
+
+    result += Math.abs(leftExcess) + Math.abs(rightExcess);
+
+    return totalExcess;
+  }
+};