源式羽语

class Solution(object):     def findMin(self, nums):         """         :type nums: List[int]         :rtype: int         """         l,r=0,len(nums)-1         while l             mid=l+(r-l)

# Definition for singly-linked list. # class ListNode(object): #     def __init__(self, x): #         self.val = x #         self.next = None # Definition for a binary tree node. # class Tre

# Definition for a binary tree node. # class TreeNode(object): #     def __init__(self, x): #         self.val = x #         self.left = None #         self.right = None class Solution(objec

# Definition for a binary tree node. # class TreeNode(object): #     def __init__(self, x): #         self.val = x #         self.left = None #         self.right = None class Solution(objec

class Solution(object):     def findLHS(self, nums):         """         :type nums: List[int]         :rtype: int         """         count = collections.Counter(nums)         ans = 0        

# Definition for a binary tree node. # class TreeNode(object): #     def __init__(self, x): #         self.val = x #         self.left = None #         self.right = None class Solution(objec

class Solution(object):     def isBalanced(self, root):         """         :type root: TreeNode         :rtype: bool         """         if not root:             return True         left = se

# Definition for a binary tree node. # class TreeNode(object): #     def __init__(self, x): #         self.val = x #         self.left = None #         self.right = None class Solution(objec

class Solution(object):     def hasPathSum(self, root, sum):         """         :type root: TreeNode         :type sum: int         :rtype: bool         """         if not root:             r

用python自定义树，先定义节点，然后在把节点合成树。class TreeNode(object): def __init__(self,root,left=None,right=None): self.root = root self.left = left self.right = right A = TreeNode('A') B =...

class Solution(object):     def findPeakElement(self, nums):         """         :type nums: List[int]         :rtype: int         """         l=0         r=len(nums)-1         while l      

# Definition for a binary tree node. # class TreeNode(object): #     def __init__(self, x): #         self.val = x #         self.left = None #         self.right = None class Solution(objec

class Solution(object):     def findMin(self, nums):         """         :type nums: List[int]         :rtype: int         """         l,r=0,len(nums)-1         while l=nums[r]:             mi

这个算法易于理解但是效果不是很好 class Solution(object):     def twoSum(self, numbers, target):         """         :type numbers: List[int]         :type target: int         :rtype: List[int]         """

1.二叉树遍历方式：先序遍历，后续遍历，中序遍历，层次遍历 先  if(BT): getRoot left right 中 if (BT) left get root right 后 if(BT): left right getroot 每个节点遇到三次，先序是第一次取数字，中序点二次取，后续点三次取。路径一样，出口一样，入口一样

# Definition for a binary tree node. # class TreeNode(object): #     def __init__(self, x): #         self.val = x #         self.left = None #         self.right = None class Solution(objec

# Definition for a binary tree node. # class TreeNode(object): #     def __init__(self, x): #         self.val = x #         self.left = None #         self.right = None class Solution(object):

class Solution(object):     def search(self, nums, target):         """         :type nums: List[int]         :type target: int         :rtype: bool         """         if not nums:          

class Solution(object):     def combine(self, n, k):         """         :type n: int         :type k: int         :rtype: List[List[int]]         candidates=[x for x in range(1,n+1)]         i

class Solution(object):     def searchMatrix(self, matrix, target):         """         :type matrix: List[List[int]]         :type target: int         :rtype: bool         """         if not m

class Solution(object):     def setZeroes(self, matrix):         """         :type matrix: List[List[int]]         :rtype: void Do not return anything, modify matrix in-place instead.         """

本质上是压栈出栈的过程1.写一个主题函数主题函数中要给定默认的树，给定的值，还要一个用来存路径的列表，再加一个记录路径长度的变量2.写一个找路径的函数首先要确定给的节点是不是一个空的，如果是空的则直接退出；其次，把当前节点加入到路径，把当前节点值加进路径长度然后，判断当前是不是叶子（叶子用一个变量表示）如果是叶子，则再判断叶子路径和和给定值是不是一样，一样则打印路径，不一样则不管如果不是叶子，则分...