Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example: Given the below binary tree and sum = 22,
5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1
[ [5,4,11,2], [5,8,4,5] ]
The idea is to perform a preorder traversal from the root node, once we found a leaf node and that leaf node's value is equal to the current sum value, then we found a solution, put it into the final result array.
The tricky part is to backtrack, think of it this way, let's say the sum is 8, and we are at node 4, see the tree below:
1 / \ 2 3 / \ 4 5
- the current solution array is [1, 2, 4], as 4 is the leaf node, we are done with this path, since it's preorder,
- The next node we need to scan is 5, and the solution list should look like [1, 2, 5] which represents the correct path,
- That's why we need to remove 4 from the solution array, otherwise, the solution array will be [1, 2, 4, 5].
Solution: 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!