应用题-二叉树专题-路径总和

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```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 ) ```
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