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Path Sum II

Problem statement

Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references.

A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22Output: [[5,4,11,2],[5,8,4,5]]Explanation: There are two paths whose sum equals targetSum:5 + 4 + 11 + 2 = 225 + 8 + 4 + 5 = 22

Example 2:

Input: root = [1,2,3], targetSum = 5Output: []

Example 3:

Input: root = [1,2], targetSum = 0Output: []

Constraints:

  • The number of nodes in the tree is in the range [0, 5000].
  • -1000 <= Node.val <= 1000
  • -1000 <= targetSum <= 1000

My solution

/**
* 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
* @param {number} targetSum
* @return {number[][]}
*/
var pathSum = function(root, targetSum) {
const result = [];

function dfs(node, path = []) {
// base case
if (!node) {
return
}
if (!node.left && !node.right) {
// console.log(node, path)
// leaf node
path.push(node.val)
if (sum(path) === targetSum) {
result.push(path)
}
// Exit if this is the lead node
return;
}

path.push(node.val)

if (node.left) {
dfs(node.left, [...path])
}

if (node.right) {
dfs(node.right, [...path])
}
}

function sum(arr) {
return arr.reduce((acc, curr) => {
acc += curr
return acc;
}, 0)
}

dfs(root);

return result;
};