随笔分类 - LeetCode
刷过的题记录一下，思路仅供参考，并非最优解。
摘要：class Solution(object): def levelOrder(self, root): """ :type root: Node :rtype: List[List[int]] """ if not root: return [] ans = [] self.dfs(1, root,
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摘要：class Solution(object): def sumOfLeftLeaves(self, root): """ :type root: TreeNode :rtype: int """ if not root or (not root.left and not root.right): r
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摘要：类似题：https://www.cnblogs.com/panweiwei/p/13585987.html class Solution(object): def lowestCommonAncestor(self, root, p, q): """ :type root: TreeNode :ty
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摘要：class Solution(object): def invertTree(self, root): """ :type root: TreeNode :rtype: TreeNode """ if not root: return root.left, root.right = root.rig
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摘要：class Solution(object): # 递归思路： # （1）如果二叉树为空，节点个数为0 # （2）如果二叉树不为空，二叉树节点个数 = 左子树节点个数 + 右子树节点个数 + 1 def countNodes(self, root): """ :type root: TreeNode
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摘要：方法一：DFS class Solution(object): # 方法一：深搜 def rightSideView(self, root): """ :type root: TreeNode :rtype: List[int] """ ans = [] self.dfs(root, 0, ans)
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摘要：class BSTIterator(object): def __init__(self, root): """ :type root: TreeNode """ self.item = [] while root: self.item.append(root) root = root.left d
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摘要：方法一：迭代 class Solution(object): # 迭代 def postorderTraversal(self, root): """ :type root: TreeNode :rtype: List[int] """ if not root: return [] ans = []
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摘要：方法一：迭代代码一： class Solution(object): # 迭代 def preorderTraversal(self, root): """ :type root: TreeNode :rtype: List[int] """ if not root: return [] pre_s
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摘要：class Solution(object): def sumNumbers(self, root): """ :type root: TreeNode :rtype: int """ if not root: return 0 return self.dfs(root, 0) def dfs(se
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摘要：class Solution(object): def flatten(self, root): """ :type root: TreeNode :rtype: None Do not return anything, modify root in-place instead. """ if no
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摘要：方法一 class Solution(object): def pathSum(self, root, sum): """ :type root: TreeNode :type sum: int :rtype: List[List[int]] """ ans = [] if not root: re
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摘要：class Solution(object): def hasPathSum(self, root, sumt): """ :type root: TreeNode :type sumt: int :rtype: bool """ if not root and sum: return False
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摘要：方法一：自底向上 class Solution(object): # 法一：自底向上 def isBalanced(self, root): """ :type root: TreeNode :rtype: bool """ return self.helper(root) >= 0 def hel
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摘要：Python3代码 class Solution: def sortedArrayToBST(self, nums: List[int]) -> TreeNode: if not nums: return None mid = len(nums) // 2 root = TreeNode(nums[
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摘要：class Solution(object): def levelOrderBottom(self, root): """ :type root: TreeNode :rtype: List[List[int]] """ if not root: return [] level_queue = []
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摘要：class Solution(object): def buildTree(self, inorder, postorder): """ :type inorder: List[int] :type postorder: List[int] :rtype: TreeNode """ # 用字典存储中
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摘要：class Solution(object): def buildTree(self, preorder, inorder): """ :type preorder: List[int] :type inorder: List[int] :rtype: TreeNode """ self.prein
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摘要：# Definition for a binary tree node. class TreeNode(object): def __init__(self, x): self.val = x self.left = None self.right = None class Solution(obj
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摘要：# Definition for a binary tree node. class TreeNode(object): def __init__(self, x): self.val = x self.left = None self.right = None class Solution(obj
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