## Branch Sums

Write a function that takes in a Binary Tree and returns a list of its branch sums ordered from leftmost branch sum to rightmost branch sum.

A branch sum is the sum of all values in a Binary Tree branch. A Binary Tree branch is a path of nodes in a tree that starts at the root node and ends at any leaf node.

Each BinaryTree node has an integer value, a left child node, and a right child node. Children nodes can either be BinaryTree nodes themselves or None / null.

### Sample Input

`1tree = 12 / \3 2 34 / \ / \5 4 5 6 76 / \ /7 8 9 10`

### Sample Output

`1[15, 16, 18, 10, 11]2// 15 == 1 + 2 + 4 + 83// 16 == 1 + 2 + 4 + 94// 18 == 1 + 2 + 5 + 105// 10 == 1 + 3 + 66// 11 == 1 + 3 + 7`

### Hints

**Hint 1**

Try traversing the Binary Tree in a depth-first-search-like fashion.

**Hint 2**

Recursively traverse the Binary Tree in a depth-first-search-like fashion, and pass a running sum of the values of every previously-visited node to each node that you’re traversing.

**Hint 3**

As you recursively traverse the tree, if you reach a leaf node (a node with no “left” or “right” Binary Tree nodes), add the relevant running sum that you’ve calculated to a list of sums (which you’ll also have to pass to the recursive function). If you reach a node that isn’t a leaf node, keep recursively traversing its children nodes, passing the correctly updated running sum to them.

**Optimal Space & Time Complexity**

O(n) time | O(n) space - where n is the number of nodes in the Binary Tree

```
1// This is the class of the input root.2// Do not edit it.3class BinaryTree {4 constructor(value) {5 this.value = value;6 this.left = null;7 this.right = null;8 }9}10
11function branchSums(root) {12 const sums = []13 findSumOfNode(root, 0, sums)14 return sums15}16
17function findSumOfNode(node, runningSum, sums) {18 // create a array of nodes19 if (!node) return;20 21 const newRunningSum = runningSum + node.value;22 if(!node.left && !node.right) {23 sums.push(newRunningSum);24 return;25 }26 27 findSumOfNode(node.left, newRunningSum, sums);28 findSumOfNode(node.right, newRunningSum, sums);29 30}
```

🧤