思路很简单 dfs 然后检查是否有一条路径返回找到了就ok. 代码如下:
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| class Solution:
def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:
if root is None:
return False
if root.left is None and root.right is None and targetSum - root.val == 0:
return True
return self.hasPathSum(root.left, targetSum - root.val) or \
self.hasPathSum(root.right, targetSum - root.val)
|