题解 | #二叉树中和为某一值的路径(一)#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param root TreeNode类 
# @param sum int整型 
# @return bool布尔型
#
class Solution:
    def hasPathSum(self , root: TreeNode, sum: int) -> bool:
        # write code here
        if not root:
            return False
        def CaculateSum(root,sum):
            if not root:
                return False
            #sum+=root.val
        #计算所有根节点到叶子节点的和，然后和sum比较
            if not root.left and not root.right:
                if root.val==sum:
                    return True

            return CaculateSum(root.left,sum-root.val) or CaculateSum(root.right,sum-root.val)
        return CaculateSum(root,sum)
        #for x in L:
         #   if x.isinstance()