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 DFS
        if not root:
            return False
        if not root.left and not root.right and sum-root.val==0:
            return True
        return self.hasPathSum(root.left,sum-root.val) or self.hasPathSum(root.right,sum-root.val)
        # write code here BFS
        '''s = []
        s += [(root, root.val)]
        while len(s):
            temp = s.pop()
            if (not temp[0].left) and (not temp[0].right) and (temp[1]==sum):
                return True
            if temp[0].left:
                s += [(temp[0].left,temp[1]+temp[0].left.val)]
            if temp[0].right:
                s += [(temp[0].right,temp[1]+temp[0].right.val)]
        return False'''