# Branch Sums

Learn how to calculate the sums of all branches in a binary tree, a fundamental task in algorithmic interviews and data structure analysis. BT
easy

Published At

7/3/2021

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

```js```tree =     1
/     \
2       3
/   \    /  \
4     5  6    7
/   \  /
8    9 10``````
```js```tree =     1
/     \
2       3
/   \    /  \
4     5  6    7
/   \  /
8    9 10``````

### Sample Output

```js```[15, 16, 18, 10, 11]
// 15 == 1 + 2 + 4 + 8
// 16 == 1 + 2 + 4 + 9
// 18 == 1 + 2 + 5 + 10
// 10 == 1 + 3 + 6
// 11 == 1 + 3 + 7``````
```js```[15, 16, 18, 10, 11]
// 15 == 1 + 2 + 4 + 8
// 16 == 1 + 2 + 4 + 9
// 18 == 1 + 2 + 5 + 10
// 10 == 1 + 3 + 6
// 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

Solution-1
```js```// This is the class of the input root.
// Do not edit it.
class BinaryTree {
constructor(value) {
this.value = value;
this.left = null;
this.right = null;
}
}

function branchSums(root) {
const sums = []
findSumOfNode(root, 0, sums)
return sums
}

function findSumOfNode(node, runningSum, sums) {
// create a array of nodes
if (!node) return;

const newRunningSum = runningSum + node.value;
if(!node.left && !node.right) {
sums.push(newRunningSum);
return;
}

findSumOfNode(node.left, newRunningSum, sums);
findSumOfNode(node.right, newRunningSum, sums);

}``````
Solution-1
```js```// This is the class of the input root.
// Do not edit it.
class BinaryTree {
constructor(value) {
this.value = value;
this.left = null;
this.right = null;
}
}

function branchSums(root) {
const sums = []
findSumOfNode(root, 0, sums)
return sums
}

function findSumOfNode(node, runningSum, sums) {
// create a array of nodes
if (!node) return;

const newRunningSum = runningSum + node.value;
if(!node.left && !node.right) {
sums.push(newRunningSum);
return;
}

findSumOfNode(node.left, newRunningSum, sums);
findSumOfNode(node.right, newRunningSum, sums);

}``````

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Have a wonderful day.
Abhishek 🙏

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