hasPathSum.py

"""
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Given the below binary tree and sum = 22,

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

"""

# 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: TreeNode, sum: int) -> bool:
        if not root:
            return False
        if not root.left and not root.right and sum == root.val:
            return True
        return self.hasPathSum(root.left,(sum-root.val)) or self.hasPathSum(root.right,(sum-root.val))

# Time Complexity = O(n) where n is the number of nodes in the tree
# Space Complexity = O(n)