112 Path Sum

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

Example 1:

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

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

Example 2:

    1
   / \
  2   3

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

Example 3:

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

Constraints:

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

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# 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 hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
        if not root :
            return False
        remain_val = targetSum - root.val
        if not root.left and not root.right:
            return remain_val == 0
        return self.hasPathSum(root.left, remain_val) or self.hasPathSum(root.right, remain_val)