Add solution #107

Author: JoshCrozier

README.md

@@ -112,6 +112,7 @@
 104|[Maximum Depth of Binary Tree](./0104-maximum-depth-of-binary-tree.js)|Easy|
 105|[Construct Binary Tree from Preorder and Inorder Traversal](./0105-construct-binary-tree-from-preorder-and-inorder-traversal.js)|Medium|
 106|[Construct Binary Tree from Inorder and Postorder Traversal](./0106-construct-binary-tree-from-inorder-and-postorder-traversal.js)|Medium|
+107|[Binary Tree Level Order Traversal II](./0107-binary-tree-level-order-traversal-ii.js)|Medium|
 108|[Convert Sorted Array to Binary Search Tree](./0108-convert-sorted-array-to-binary-search-tree.js)|Easy|
 110|[Balanced Binary Tree](./0110-balanced-binary-tree.js)|Easy|
 111|[Minimum Depth of Binary Tree](./0111-minimum-depth-of-binary-tree.js)|Easy|

solutions/0107-binary-tree-level-order-traversal-ii.js

@@ -0,0 +1,44 @@
+/**
+ * 107. Binary Tree Level Order Traversal II
+ * https://leetcode.com/problems/binary-tree-level-order-traversal-ii/
+ * Difficulty: Medium
+ *
+ * Given the root of a binary tree, return the bottom-up level order traversal of its nodes' values.
+ * (i.e., from left to right, level by level from leaf to root).
+ */
+
+/**
+ * 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 levelOrderBottom = function(root) {
+  if (!root) return [];
+
+  const result = [];
+  const stack = [root];
+
+  while (stack.length > 0) {
+    const level = [];
+    const n = stack.length;
+
+    for (let i = 0; i < n; i++) {
+      const node = stack.shift();
+      level.push(node.val);
+
+      if (node.left) stack.push(node.left);
+      if (node.right) stack.push(node.right);
+    }
+
+    result.unshift(level);
+  }
+
+  return result;
+};