hasPathSum.py

  # Path Sum

# Solution
# 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.

# Example:

# Given the below binary tree and sum = 22,

      # 5
     # / \
    # 4   8
   # /   / \
  # 11  13  4
 # /  \      \
# 7    2      1
# return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

# 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 root.val == sum:
            return True
        
        sum -= root.val
        
        return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)