# Javascript Solution For Path Sum IV - Sum Of All Paths From The Root Towards The Leaves

**Path Sum IV** is an example of another **tree** problems. In this post we will see how we can solve in Javascript.

## Problem Description

If the depth of a tree is smaller than 5, then this tree can be represented by a list of three-digits integers.

For each integer in this list:

- The hundreds digit represents the depth D of this node, 1 <= D <= 4.
- The tens digit represents the position P of this node in the level it belongs to, 1 <= P <= 8. The position is the same as that in a full binary tree.
- The units digit represents the value V of this node, 0 <= V <= 9.

Given a list of ascending three-digits integers representing a binary with the depth smaller than 5. You need to return the sum of all paths from the root towards the leaves.

## Example 1

Input: [113, 215, 221] Output: 12

Explanation: The tree that the list represents is:

3 / \ 5 1

The path sum is (3 + 5) + (3 + 1) = 12.

## Example 2

Input: [113, 221] Output: 4

Explanation: The tree that the list represents is:

3 \ 1

The path sum is (3 + 1) = 4.

## Solution

How do we solve problem like this if we were given a normal tree?

Yes, traverse it, keep a root to leaf running sum. If we see a leaf node (node.left == null && node.right == null), we add the running sum to the final result.

Now each tree node is represented by a number. 1st digits is the level, 2nd is the position in that level (note that it starts from 1 instead of 0). 3rd digit is the value. We need to find a way to traverse this tree and get the sum.

The idea is, we can form a tree using a HashMap. The key is first two digits which marks the position of a node in the tree. The value is value of that node. Thus, we can easily find a node's left and right children using math.

**Formula: **For node xy? its left child is (x+1)(y*2-1)? and right child is (x+1)(y*2)?

Given above HashMap and formula, we can traverse the tree. Problem is solved!

**Implementation: **Please check the main.js snippet for the solution. If you have different approach in mind or have any suggestion for this implementation feel free to share in the comment below. Thanks!

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