path-sum-ii

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Description

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 = 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

Solution(Python)

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
        paths = []
        path = []

        def preorder(curr_node, sum_so_far):
            nonlocal path
            if not curr_node:
                return
            sum_so_far += curr_node.val
            path.append(curr_node.val)

            if not curr_node.left and not curr_node.right:
                if sum_so_far == targetSum:
                    paths.append(path[:])
            if curr_node.left:
                preorder(curr_node.left, sum_so_far)
            if curr_node.right:
                preorder(curr_node.right, sum_so_far)

            path.pop()

        preorder(root, 0)
        return paths