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
#
# 每次进入左子树和右子树，在叶子节点进行判断，只要有一个满足即可
class Solution:
    def hasPathSum(self , root: TreeNode, sum: int) -> bool:
        if not root:
            return False
            
        def dfs(r, target):
            target -= r.val

            if not r.left and not r.right:
                return target == 0
            elif r.left and not r.right:
                return dfs(r.left, target)
            elif not r.left and r.right:
                return dfs(r.right, target)
            else:
                return dfs(r.left, target) or dfs(r.right, target)
        
        return dfs(root, sum)