力扣日刷48

102.二叉树的层序遍历我们需要一层又一层地去遍历二叉树具体来说我们cur数组去装当前的值我们用nxt数组去装载下一层的值当cur全部为空的时候退出装载。下面就是具体的代码有一点困惑的地方就是该怎么生成nxt。其他的我觉得没什么问题。# Definition for a binary tree node. # class TreeNode: # def __init__(self, val0, leftNone, rightNone): # self.val val # self.left left # self.right right class Solution: def levelOrder(self, root: Optional[TreeNode]) - List[List[int]]: if root is None: return [] ans [] cur [root] while cur: nxt [] vals [] for node in cur: vals.append(node.val) if node.left: nxt.append(node.left) if node.right: nxt.append(node.right) cur nxt ans.append(vals) return ans103.二叉树的锯齿形排序方法代码与上一题相似只是需要设置一个反转判断的变量以及一个反转的判断# Definition for a binary tree node. # class TreeNode: # def __init__(self, val0, leftNone, rightNone): # self.val val # self.left left # self.right right class Solution: def zigzagLevelOrder(self, root: Optional[TreeNode]) - List[List[int]]: if root is None: return [] ans [] cur [root] even False while cur: nxt [] vals [] for node in cur: vals.append(node.val) if node.left: nxt.append(node.left) if node.right: nxt.append(node.right) cur nxt ans.append(vals[::-1] if even else vals) even not even return ans513.找到树左下角的值其实也用层序排列可以做只需要去选择出来那个值就可以了# Definition for a binary tree node. # class TreeNode: # def __init__(self, val0, leftNone, rightNone): # self.val val # self.left left # self.right right class Solution: def findBottomLeftValue(self, root: Optional[TreeNode]) - int: ans [] cur [root] while cur: vals [] nxt [] for node in cur: vals.append(node.val) if node.left: nxt.append(node.left) if node.right: nxt.append(node.right) cur nxt ans.append(vals) return ans[-1][0]17.电话号码的字母组合这里对递过去归回来的思路有了一定的了解。1.当前要做什么分为什么时候我把结果输出边界条件2.子问题是什么我不输出的时候需要干什么非边界条件3.下一个子问题是什么 递归过程这里的边界条件就是当字母的数字Map [,,abc,def,ghi,jkl,mno,pqrs,tuv,wxyz] class Solution: def letterCombinations(self, digits: str) - List[str]: n len(digits) if n 0: return [] path [] * n ans [] def dfs(i): if i n: return ans.append(.join(path)) for c in Map[int(digits[i])]: path[i] c dfs(i1) dfs(0) return ans