应用题-二叉树专题-路径总和
```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 hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
def exist( r, t ):
if not r: return False
f = t - r.val
if not r.left and not r.right: return f == 0
st1 = exist( r.left, f )
if st1: return True
st2 = exist( r.right, f )
if st2: return True
return st1 or st2
return exist( root, targetSum )
```