[Leetcode]回溯算法——python版本

本篇文章根据labuladong的算法小抄汇总回溯算法的相关常见算法，采用python3实现

回溯算法框架（DFS）

• 回溯算法就是DFS算法（depth first searc，深度优先搜索算法），本质上是一种暴力穷举算法

• 回溯问题实际上就是决策树的遍历过程：

  1、路径：已经做出的选择

  2、选择列表：当前可以做的选择

  3、结束条件：到达决策树底层，无法再做选择的条件

• 回溯算法的框架

result = []
def backtrack(路径,选择列表):
    if 满足结束条件:
        result.add(路径)
        return
    for 选择 in 选择列表:
        做选择
        backtrack(路径，选择列表)
        撤销选择

全排列问题

问题：n个不重复的数，全排列

import copy
res = []

#主函数，输入一组不重复的数字，返回它们的全排列
def permute(nums):
    track = []    #记录路径
    backtrack(nums,track)
    return res

def backtrack(nums,track):
    #触发结束条件
    if len(track) == len(nums):
        track_ = copy.copy(track)
        res.append(track_)
        return
    for i in range(len(nums)):
        #排除不合法选择
        if nums[i] in track:
            continue
        #做选择
        track.append(nums[i])
        #进入下一层决策树
        backtrack(nums,track)
        #取消选择
        track.pop()

• 回溯算法不像动态规划存在重叠子问题可以优化，它就是纯暴力穷举，复杂度一般都很高

N皇后问题

问题：给定一个N*N的棋盘，让你放置N个皇后，使得它们不能互相攻击（皇后可以共计同一行、同一列、左上左下右上右下四个方向的任意单位）

class Solution:
    def solveNQueens(self,n):
        def backtrack(board,row):
            if len(board) == row:
                res.append(copy.deepcopy(board))
            for col in range(len(board)):
                if not self.isValid(board,row,col):
                    continue
                board[row][col] = 'Q'
                backtrack(board,row+1)
                board[row][col] = '.'

        res = []
        board = [['.' for i in range(n)] for j in range(n)]
        backtrack(board,0)
        for i in range(res):
            for j in range(n):
                res[i][j] = "".join(res[i][j])
        return res
	def isValid(self,board,row,col):
        n = len(board)
        
        for i in range(row):
            if row[i][col] == "Q":
                return False
            
        i = row - 1
        j = col - 1
        while (i >= 0) and (j >= 0):
            if board[i][j] == "Q":
                return False
            i -= 1
            j -= 1
        
        i = row - 1
        j = col + 1
        while (i >= 0) and (j < n):
            if board[i][j] == "Q":
                return False
            i -= 1
            j += 1
        return True

划分为k个相等子集

#数组视角，O(k^n)
def canPartitionKSubsets(nums,k):
    if k > len(nums):
        return False
    sumNum = sum(nums)
    if sumNum % k != 0:
        return False
    target = sumNum // k
    bucket = [0 for i in range(k)]
    nums.sort(reverse=True)
    return backtrack(nums,0,bucket,target)
def backtrack(nums,index,bucket,target):
    if index == len(nums):
        for i in range(len(bucket)):
            if bucket[i] != tarket:
                return False
        return True
    for i in range(len(bucket)):
        if bucket[i] + nums[index] > target:
            continue
        bucket[i] += nums[index]
        if backtrack(nums,index+1,bucket,target):
            return True
        bucket[i] -= nums[index]
    return False

#桶视角，O(k*2^n)
def canPartitionKSubsets(nums,k):
    def backtrack(k,bucket,nums,0,used,target):
        if k == 0:
            return True
        if bucket == target:
            return backtrack(k-1,bucket,nums,0,used,target)
        for i in range(start,len(nums)):
            if used[i]:
                continue
            if nums[i] + bucket > target:
                continue
            used[i] = True
            bucket += nums[i]
            if backtrack(k,bucket,nums,i+1,used,target):
                return True
            used[i] = False
            bucket -= nums[i]
        return False
    
    if k > len(nums):
        return False
    sumNum = sum(nums)
    if sumNum % k != 0:
        return False
    used =[False for i in range(len(nums))]
    target = sumNum // k
    return backtrack(k,0,nums,0,used,target)
	
    

• 尽量少量多次，宁可二选一选k次，也不要k选1选一次。

求子集

#数学归纳，O(N*2^N)
def subsets(nums):
    if len(nums) == 0:
        return [[]]
    n = nums.pop()
    res = subsets(nums)
    for i in range(len(res)):
        x = copy.deepcopy(res[i])
        res.append(res[i])
        res[-1].append(n)
    return res

#回溯
def subsets(nums):
    def backtrack(nums,start,track):
        res.append(track[:])
        for i in range(start,len(nums)):
            track.append(nums[i])
            backtrack(nums,i+1,track)
            track.pop()
        
    track = []
    res = []
    backtrack(nums,0,track)
    return res

求组合

def combine(n,k): #输出[1,..,n中k个数字的所有组合]
    def backtrack(n,k,start,track):
        if k == len(track):
            res.append(track[:])
            return
        for i in range(start,n+1):
            track.append(i)
            backtrack(n,k,i+1,track)
            track.pop()
    
    res = []
    if (k <= 0) or (n <= 0):
        return res
    track = []
    backtrack(n,k,1,track)
    return res

解数独

def solveSudoku(board):
    
    def isValid(board,r,c,n):
        for i in range(9):
            if board[i][c] == n:
                return False
            if board[r][i] == n:
                return False
            if board[(r//3)*3 + i//3][(c//3)*3 + i%3] == n:
                return False
        return True
    
    def backtrack(board,i,j):
        m = 9
        n = 9
        if i == m:
            return True
        if j == n:
            return backtrack(board,i+1,0)
        if board[i][j] != ".":
            return backtrack(board,i,j+1)
        for num in range(1,10):
            if not isValid(board,i,j,str(num)):
                continue
            board[i][j] = str(num)
            if backtrack(board,i,j+1):
                return True
            board[i][j] = "."
        return False
        
    backtrack(board,0,0)

括号生成

合法括号的性质：

• 左括号数量一定等于右括号数量
• 对于一个合法的括号字符串组合p，必然对于任何0 <= i < len§都有p[0…i]中左括号数量大于等于右括号数量

def generateParenthesis(n):
    def backtrack(left,right,track,res):
        if right < left:
            return
        if (left < 0) or (right < 0):
            return
        if (left == 0) and (right == 0):
            res.append(track[:])
            return
        track.append("(")
        backtrack(left - 1,right,track,res)
        track.pop()
        
        track.append(")")
        backtrack(left,right-1,track,res)
        track.pop()
    
    if n == 0:
        return []
    res = []
    track=[]
    backtrack(n,n,track,res)
    return res

BFS算法框架

• 核心思想：把一些问题抽象成图，从一个点开始，向四周开始扩散
• BFS找到的路径一定最短，但空间复杂度可能比DFS大很多
• 常见场景：在一幅【图】中找到从起点start到终点target的最近距离

#计算从起点到终点的最近距离
def BFS(start,target):
    q = [] #核心数据结构
    q.append(start) #将起点加入队列
    visited.append(start) #避免走回头路
    step = 0 #记录扩散的步数
    while len(q) != 0:
        sz = len(q)
        for i in range(sz):
            cur

二叉树的最小高度

打开密码锁的最小步数

数据结构

二叉树

翻转二叉树

class TreeNode:
    def __init__(self,val=0,left=None,right=None):
        self.val = val
        self.left = left
        self.right = right
class Solution:
    def invertTree(self,root):
        if root is None:
            return None
        tmp = root.left
        root.left = root.right
        root.right = tmp
        self.invertTree(root.left)
        self.invertTree(root.right)
        return root

填充二叉树节点的右侧指针

def connect(root):
    if root is None:
        return None
    connectTwoNode(root.left,root.right)
    return root
def connectTwoNode(node1,node2):
    if (node1 is None) or (node2 is None):
        return
    node1.next = node2
    connectTwoNode(node1.left,node1.right)
    connectTwoNode(node2.left,node2.right)
    connectTwoNode(node1.right,node2.left)

二叉树展开为链表

def flatten(root):
    if root is None:
        return None
    flatten(root.left)
    flatten(root.right)
    left = root.left
    right = root.right
    root.left = None
    root.right = left
    p = root
    while p.right is not None:
        p = p.right
    p.right = right

构造最大二叉树

#超过递归深度
def constructMaximumBinaryTree(nums):
    return build(nums, 0, nums.length - 1)
def build(nums, lo , hi):
    if lo > hi:
        return None
    index = -1
    maxVal = float("-inf")
    for i in range(lo,hi+1):
        if maxVal < nums[i]:
            maxVal = nums[i]
            index = i
            
    root = TreeNode(maxVal)
    root.left = build(nums,lo,index-1)
    root.right = build(nums,index+1,hi)
    return root

通过前序和中序遍历结果构造二叉树

def buildTree(preorder,inorder):
    return build(preorder,0,len(preorder)-1,inorder,0,len(inorder)-1)
def build(preorder,preStart,preEnd,inorder,inStart,inEnd):
    if preStart > preEnd:
        return None
    rootVal = preorder[preStart]
    index = 0
    for i in range(inStart,inEnd+1):
        if inorder[i] == rootVal:
            index = i
            break
    root = TreeNode(rootVal)
    leftSize = index - inStart
    root.left = build(preorder,preStart+1,preStart+leftSize,inorder,inStart,index-1)
    root.right = build(preorder,preStart+leftSize+1,preEnd,inorder,index+1,inEnd)
    return root

通过中序和后序遍历结果构造二叉树

class TreeNode:
    def __init__(self,val=0,left=None,right=None):
        self.val = val
        self.left = left
        self.right = right
def buildTree(inorder,postorder):
    def build(inorder,instart,inend,postorder,poststart,postend):
        if instart > inend:
            return None
        rootValue = postorder[postend]
        for i in range(instart,inend+1):
            if inorder[i] == rootValue:
                index = i
                break
        root = TreeNode(rootValue)
        leftSize = index-instart
        root.left = build(inorder,instart,index-1,postorder,poststart,poststart+leftSize-1)
        root.right = build(inorder,index+1,inend,postorder,poststart+leftSize,postend-1)
        return root
    return build(inorder,0,len(inorder)-1,postorder,0,len(postorder)-1)

寻找重复子树

class treeNode():
    def __init__(self,val=0,left=None,right=None):
        self.val = val
        self.left = left
        self.right = right
class Solution:
    def findDuplicateSubtrees(self, root: Optional[TreeNode]) -> ist[Optional[TreeNode]]:
        def traverse(root):
            if root is None:
                return None
            left = traverse(root.left)
            right = traverse(root.right)
            subTree = str(left) + "," + str(right)+ "," + str(root.val)
            fre = subTree_fre.setdefault(subTree,0)
            if fre == 1:
                res.append(root)
            subTree_fre[subTree] = fre + 1
            return subTree
        res = []
        subTree_fre = dict()
        traverse(root)
        return res

二叉树的序列化与反序列化

前序遍历

class Codec:
    def serialize(self,root):
        if root is None:
            return "None"
        return str(root.val)+","+str(self.serialize(root.left))+","+str(self.serialize(root.right))
    def deserialize(self,data):
        def dfs(datalist):
            data = datalist.pop(0)
            if data == "None":
                return None
            root = TreeNode(int(data))
            root.left = dfs(datalist)
            root.right = dfs(datalist)
            return root
        datalist = data.split(",")
        return dfs(datalist)

• 中序遍历无法实现反序列化

层级遍历

class Codec:
    def serialize(self,root):
        if root is None:
            return ""
        queue = collections.deque([root])
        res = []
        while queue:
            node = queue.popleft()
            if node is None:
                res.append("None")
               	continue
            res.append(str(node.val))
            queue.append(node.left)
            queue.append(node.right)
        return ",".join(res)
    def deserialize(self,data):
        if not data:
            return []
        dataList = data.split(",")
        root = TreeNode(int(dataList[0]))
        queue = collections.deque(root)
        i = 1
        while queue:
            node = queue.popleft()
            if dataList[i] != "None":
                node.left = TreeNode(int(dataList[i]))
                queue.append(node.left)
            i += 1
            if dataList[i] != "None":
                node.right = TreeNode(int(data))
                queue.append(node.right)
            i += 1
        return root
        
                

扁平化嵌套列表迭代器

class NestedIterator:
    def __init__(self,nestedList):
        self.q = collections.deque()
        self.dfs(nestedList)
    def dfs(self,nestedList):
        for elem in nestedList:
            if elem.isInteger():
                self.q.append(elem.getInteger())
            else:
                self.dfs(elem.getList())
    def next(self):
        return self.q.popleft()
    def hasNext(self):
        return len(self.q)

二叉树的最近公共祖先LCA

def lowestCommonAncestor(root,p,q):
    if root is None:
        return None
    if (root == p) or (root == q):
        return root
    left = lowestCommonAncestor(root.left,p,q)
    right = lowestCommonAncestor(root.right,p,q)
    if (left is not None) and (right is not None):
        return root
    if (left is None) and (right is None):
        return None
    return right if left is None else left

完全二叉树的节点个数

完全二叉树：

满二叉树：

#普通二叉树，时间复杂度O(N)
def countNodes(root):
    if root is None:
        return 0
    return 1 + countNodes(root.left) + countNodes(root.right)
#满二叉树，节点总数和树的高度呈指数关系
def countNodes(root):
    h = 0
    while root:
        root = root.left
        h += 1
    return pow(2,h) - 1
#完全二叉树,时间复杂度O(logN*logN)
def countNodes(root):
    if root is None:
        return 0
    l = root
    r = root
    hl = 0
    hr = 0
    while l:
        l = l.left
        hl += 1
    while r:
        r = r.right
        hr += 1
    if hl == hr:
        return pow(2,hl) - 1
    return 1 + countNodes(root.left) + countNodes(root.right)

• 由于完全二叉树的性质，其子树一定有一棵是满的

递归改迭代

stk = collections.deque()
#左侧遍历到底，存入栈
def pushLeftBranch(p):
    while p:
        #前序遍历代码位置
        stk.append(p)
        p = p.left
def traverse(root):
    visited = TreeNode(-1) #指向上一次遍历完的子树根节点
    pushLeftBranch(root)
    while stk:
        p = stk[-1]
        if ((p.left is None) or (p.left == visited)) and (p.right != visited):
            #中序遍历代码位置
            pushLeftBranch(p.right)
        if (p.right is None) or (p.right == visited):
            #后序遍历代码位置
            visited = stk.pop(-1)

• 除了BFS层级遍历外，二叉树还是用递归，因为递归最符合二叉树结构他点

二叉搜索树(BST)

• BST特点：

  1、对于BST的每个节点node，左子树节点的值 < node的值 < 右子树节点的值

  2、对于BST的每个节点node，左子树和右子树都是BST

  3、BST的中序遍历结果是有序的（升序）

寻找第K小的元素

class Solution:
    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
        self.res = 0
        self.rank = 0
        self.traverse(root,k)
        return self.res
    
    def traverse(self,root,k):
        if root is None:
            return None
        self.traverse(root.left,k)
        self.rank += 1
        if self.rank == k:
            self.res = root.val
            return
        self.traverse(root.right,k)

BST转化累加树

class Solution():
    def convertTree(root):
        def convert(root):
            if root is None:
                return None
            convert(root.right)
            self.sum += root.val
            root.val = self.sum
            convert(root.left)
        self.sum = 0
        convert(root)
        return root

判断BST合法性

def isValidBST(root):
    return isValidBST(root,None,None)
def isValidBST(root,min,max):
    if root is None:
        return True
    if min is not None and root.val <= min.val:
        return False
    if max is not None and root.val >= max.val:
        return False
    return isValidBST(root.left,min,root) and isValidBST(root.right,root,max)    

在BST中搜索元素

def search(root,val):
    if root is None:
        return None
    if val > root.val:
        return search(root.right,val)
    if val < root.val:
        return search(root.left,val)
    return root

在BST中插入一个数

def insertIntoBST(root,val):
    if root is None:
        return TreeNode(val)
    if val > root.val:
        root.right = insertIntoBST(root.right,val)
    if val < root.val:
        root.left = insertIntoBST(root.left,val)
    return root

在BST中删除一个数

class Solution:
    def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
        def getMin(root):
            while root.left is not None:
                root = root.left
            return root
        if root is None:
            return None
        if root.val == key:
            if root.left is None:
                return root.right
            elif root.right is None:
                return root.left
            minNode = getMin(root.right)
            minNode.left = root.left
            minNode.right = root.right
            root = minNode
        elif root.val < key:
            self.deleteNode(root.right,key)
        else:
            self.deleteNode(root.left,key)
        return root

不同的二叉搜索树

class Solution:
    def numTrees(self, n: int) -> int:
        memo = [[0 for i in range(n+1)] for j in range(n+1)]
        return count(1,n,memo)
def count(lo,hi,memo):
    if lo > hi:
        return 1
    if memo[lo][hi] != 0:
        return memo[lo][hi]
    res = 0
    for i in range(lo,hi+1):
        left = count(lo,i-1,memo)
        right = count(i+1,hi,memo)
        res += left * right
    memo[lo][hi] = res
    return res

不同的二叉搜索树②

def generateTrees(n):
    if n == 0:
        return TreeNode()
    return build(1,n)
def build(lo,hi):
    res = []
    if lo > hi:
        return r[None]
    for i in range(lo,hi+1):
        leftTree = build(lo,i-1)
        rightTree = build(i+1,hi)
        for left in leftTree:
            for right in rightTree:
                root = TreeNode(i)
                root.left = left
                root.right = right
                res.append(root)
    return res

二叉搜索子树的最大键值和

import sys
sum = 0
def maxSumBST(root):
	traverse(root)
    return maxSum
def traverse(root): #返回[isBST,min,max,sum]
    if root is None:
        return [1,sys.maxsize,-sys.maxsize-1,0]
    left = traverse(root.left)
    right = traverse(root.right)
    res = [0 for i in range(4)]
    if (left[0] == 1) and (right[0] == 1) and (left[2] < root.val < right[1]):
        #以root为根的二叉树是BST
        res[0] = 1
        res[1] = min(left[1],root.val)
        res[2] = max(right[2],root.val)
        res[3] = left[3] + right[3] + root.val
        maxSum = max(maxSum,res[3])
    else:
        res[0] = 0
    return res

• BST相关问题：（1）利用BST左小右大的特性提升算法效率；（2）中序遍历的特性(有序)满足题目要求
• 如果当前节点要做的事情需要通过左右子树的计算结果推导出来，就要用到后序遍历
• 要尽可能避免递归函数中调用其他递归函数