记录了初步解题思路 以及本地实现代码；并不一定为最优 也希望大家能一起探讨 一起进步

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2/26 938. 二叉搜索树的范围和

dfs 判断节点值是否在区间内

class TreeNode(object):def __init__(self, val=0, left=None, right=None):self.val = valself.left = leftself.right = right
def rangeSumBST(root, low, high):""":type root: TreeNode:type low: int:type high: int:rtype: int"""def dfs(node):ret = 0if node.val>low:if node.left:ret = dfs(node.left)if node.val>=low and node.val<=high:ret+=node.valif node.val<high:if node.right:ret += dfs(node.right)return retreturn dfs(root)

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2/27 2867. 统计树中的合法路径数目

欧拉筛筛选质数
以质数节点为根 深搜所有非质数的子树
求字数大小 任意两个不同子树节点 路径都通过质数根节点
只有一个节点为质数 符合题意

def countPaths(n, edges):""":type n: int:type edges: List[List[int]]:rtype: int"""prime=[]isprime = [True]*(n+1)isprime[1]=Falsefor i in range(2,n+1):if isprime[i]:prime.append(i)for p in prime:if p*i>n:breakisprime[p*i]=Falseif i%p==0:breakm=[[] for _ in range(n+1)]for i,j in edges:m[i].append(j)m[j].append(i)global seendef dfs(i,pre):global seenseen.append(i)for j in m[i]:if j!=pre and not isprime[j]:dfs(j,i)ans = 0cnt=[0]*(n+1)for i in range(1,n+1):if not isprime[i]:continuecur = 0for j in m[i]:if isprime[j]:continueif cnt[j]==0:seen = []dfs(j,0)for k in seen:cnt[k] = len(seen)ans+=cnt[j]*curcur+=cnt[j]ans+=curreturn ans

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2/28 2673. 使二叉树所有路径值相等的最小代价

满二叉树有n个节点 那么存在(n+1)//2个叶子节点
对于两个兄弟叶子节点 除了改变其自身 无法改变两个节点路径和的差值
所以将小的叶子节点增加到大的即可
改完叶子节点 将叶子节点的值增加到父节点中 依旧考虑父节点和叔节点的情况

def minIncrements(n, cost):""":type n: int:type cost: List[int]:rtype: int"""ans=0for i in range(n-2,0,-2):ans += abs(cost[i]-cost[i+1])cost[i//2] += max(cost[i],cost[i+1])return ans

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2/29 2581. 统计可能的树根数目

首先计算以0位根节点时猜对的个数cnt
当根节点从0换到其某个子节点x时
除了0,x的父子关系发生变化 其他没有变化
所以如果此时(0,x)在猜测中 则cnt-1
如果(x,0)在猜测中 则cnt+1
遇到cnt>=k 则ans+1

def rootCount(edges, guesses, k):""":type edges: List[List[int]]:type guesses: List[List[int]]:type k: int:rtype: int"""g=[[] for _ in range(len(edges)+1)]for x,y in edges:g[x].append(y)g[y].append(x)s = {(x,y) for x,y in guesses}global cnt,ansans = 0cnt = 0def dfs(x,p):global cntfor y in g[x]:if y!=p:cnt += (x,y) in sdfs(y,x)dfs(0,-1)def reroot(x,p,cnt):global ansif cnt>=k:ans+=1for y in g[x]:if y!=p:reroot(y,x,cnt-((x,y) in s) +((y,x) in s))reroot(0,-1,cnt)return ans

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3/1 2369. 检查数组是否存在有效划分

dp[i+1]判断是否能够有效划分0~i

def validPartition(nums):""":type nums: List[int]:rtype: bool"""n=len(nums)dp=[False]*(n+1)dp[0]=Truefor i,x in enumerate(nums):if (i>0 and dp[i-1] and x==nums[i-1]) or (i>1 and dp[i-2] and (x==nums[i-1]==nums[i-2] or x==nums[i-1]+1==nums[i-2]+2)):dp[i+1]=Truereturn dp[n]

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3/2 2368. 受限条件下可到达节点的数目

不考虑受限的节点 建图
DFS搜索

def reachableNodes(n, edges, restricted):""":type n: int:type edges: List[List[int]]:type restricted: List[int]:rtype: int"""s = set(restricted)m = [[] for _ in range(n)]for x,y in edges:if x not in s and y not in s:m[x].append(y)m[y].append(x)def dfs(x,f):cnt = 1for y in m[x]:if y!=f:cnt += dfs(y,x)return cntreturn dfs(0,-1)

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3/3

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