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.

Leetcode link

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: [[5,4,11,2],[5,8,4,5]]

Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22

Example 2:

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

Example 3:

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

Constraints:

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

Code

Python Code

class Solution(object):
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        res = []
        self.preorder(root, [], sum, res)
        return res

    def preorder(self, node, l, left, res):
        if not node: return
        if not node.left and not node.right: # If node is a leaf
            if left == node.val: # If node.val is equal to the value left
                res.append(l + [node.val]) # It is one of the answers, save it to res[]
        else:
            self.preorder(node.left, l+[node.val], left - node.val, res)
            self.preorder(node.right, l+[node.val], left - node.val, res)