题解 | #二叉树中和为某一值的路径(一)#

python递归

class Solution:
    def hasPathSum(self , root: TreeNode, sum: int) -> bool:
        # write code here
        if not root: return False
        elif not root.left and not root.right:
            return True if root.val==sum else False
        else:
            return self.hasPathSum(root.left,sum-root.val) or self.hasPathSum(root.right,sum-root.val)