leetcode python 解题模板

一般的题

class Solution(object):
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        dic = { "2": "abc", "3": "def", "4":"ghi", "5":"jkl", "6":"mno", "7":"pqrs", "8":"tuv", "9":"wxyz"}
        
        res=[]
        if len(digits) ==0:
            return res
            
        self.dfs(digits, 0, dic, '', res)
        return res
    
    def dfs(self, nums, index, dic, path, res):
        if index >=len(nums):
            res.append(path)
            return
        string1 =dic[nums[index]]
        for i in string1:
            self.dfs(nums, index+1, dic, path + i, res)
            
solution = Solution()
print(solution.letterCombinations("23"))
#这个self可以不用传入，不管就行了

 涉及到树的题（参考：https://blog.csdn.net/qq_43355165/article/details/122780188）

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution(object):
    # 1. 递归
    def mirrorTree_1(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if not root: return
        # tmp = root.left
        # root.left = self.mirrorTree(root.right)
        # root.right = self.mirrorTree(tmp)
        root.left, root.right = self.mirrorTree(root.right), self.mirrorTree_1(root.left)
        return root
    # 2. 辅助栈
    def mirrorTree(self, root):
        if not root: return
        stack = [root]
        while stack:
            node = stack.pop()
            if node.left: stack.append(node.left)
            if node.right: stack.append(node.right)
            node.left, node.right = node.right, node.left
        return root

t1 = TreeNode(
4
)
t2 = TreeNode(
2
)
t3 = TreeNode(
7
)
t4 = TreeNode(
1
)
t5 = TreeNode(
3
)
t6 = TreeNode(
6
)
t7 = TreeNode(
9
)

t2.left = t4
t2.right = t5
t3.left = t6
t3.right = t7
t1.left = t2
t1.right = t3
solution = Solution()
a = solution.mirrorTree(t1)
print(a.__dict__)