摘要: 题目描述: 参考后的提交: class Solution(object): def pathSum(self, root, sum): """ :type root: TreeNode :type sum: int :rtype: List[List[int]] """ r = [] l = [] 阅读全文
posted @ 2019-03-27 19:18 oldby 阅读(87) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 第一次提交: class Solution(object): def hasPathSum(self, root, sum): """ :type root: TreeNode :type sum: int :rtype: bool """ if not root : return Fa 阅读全文
posted @ 2019-03-27 15:03 oldby 阅读(123) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 第一次提交: class Solution(object): def minDepth(self, root): """ :type root: TreeNode :rtype: int """ if not root: return 0 if root.left and root.ri 阅读全文
posted @ 2019-03-27 12:10 oldby 阅读(136) 评论(0) 推荐(0) 编辑
摘要: 恢复内容开始 题目描述: 方法一: class Solution(object): def isBalanced(self, root): """ :type root: TreeNode :rtype: bool """ if not root: return True if abs(self.c 阅读全文
posted @ 2019-03-26 22:04 oldby 阅读(179) 评论(0) 推荐(0) 编辑
摘要: 恢复内容开始 题目描述: 方法一: class Solution(object): def sortedArrayToBST(self, nums): """ :type nums: List[int] :rtype: TreeNode """ if not nums: return None el 阅读全文
posted @ 2019-03-25 21:08 oldby 阅读(179) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一: class Solution: def combinationSum(self, candidates, target): """ :type candidates: List[int] :type target: int :rtype: List[List[int]] """ 阅读全文
posted @ 2019-03-23 20:41 oldby 阅读(163) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一; class Solution(object): def levelOrder(self, root): if not root: return [] ans = [] stack = [root] while stack: tem_stack = [] tem_ans = [] 阅读全文
posted @ 2019-03-21 21:49 oldby 阅读(121) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一: class Solution(object): def levelOrderBottom(self, root): """ :type root: TreeNode :rtype: List[List[int]] """ if not root: return [] ans = 阅读全文
posted @ 2019-03-21 21:38 oldby 阅读(110) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 第一次提交: class Solution(object): def maxDepth(self, root): """ :type root: TreeNode :rtype: int """ if not root: return 0 elif root.left==None and 阅读全文
posted @ 2019-03-21 18:28 oldby 阅读(180) 评论(0) 推荐(0) 编辑
摘要: 题目描述: 方法一:递归: class Solution: def isSymmetric(self, root: TreeNode) -> bool: if not root: return True def Tree(p, q): if not p and not q: return True 阅读全文
posted @ 2019-03-20 14:43 oldby 阅读(237) 评论(0) 推荐(0) 编辑