代码随想录算法训练营第二十二天|LeetCode 77 组合，LeetCode 216 组合总和 III，LeetCode 450 删除二叉搜索树中的节点

1.LeetCode 77 组合

题目链接：77. 组合

class Solution:
    def combine(self, n: int, k: int) -> List[List[int]]:
        def backtracking(n, k, startIndex, path, result):
            if len(path) == k:
                result.append(path[:])
                return
            
            for i in range(startIndex, n - (k - len(path)) + 2):
                path.append(i)
                backtracking(n, k, i + 1, path, result)
                path.pop()

        result = []
        backtracking(n, k, 1, [], result)
        return result

确实不好理解

第一题结束

2.LeetCode 216 组合总和 III

题目链接：216. 组合总和 III

class Solution:
    def combinationSum3(self, k: int, n: int) -> List[List[int]]:
        def backtracking(k, n, Sum, Index, path, result):
            if Sum > n:
                return
            
            if len(path) == k:
                if Sum == n:
                    result.append(path[:])
                return

            for i in range(Index, 9 - (k - len(path)) + 2):
                Sum += i
                path.append(i)
                backtracking(k, n, Sum, i + 1, path, result)
                Sum -= i
                path.pop()
        result = []
        backtracking(k, n, 0, 1, [], result)
        return result

跟前面那题差不多

3.LeetCode 17 电话号码的字母组合

题目链接：17. 电话号码的字母组合

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        lettermap = ["",
            "",     
            "abc",  
            "def",  
            "ghi",  
            "jkl",  
            "mno",  
            "pqrs", 
            "tuv",  
            "wxyz"  ]
        result = []
        s = ''

        def backtracking(digits, Index):
            nonlocal s, result
            if Index == len(digits):
                result.append(s)
                return
            
            digit = int(digits[Index])
            letters = lettermap[digit]

            for i in range(len(letters)):
                s += letters[i]
                backtracking(digits, Index + 1)
                s = s[:-1]
        if len(digits) == 0:
            return result
        backtracking(digits, 0)
        return result

还行吧

今天用时2.5h，回溯比二叉树好理解点