樹及其衍生演算法(Trees and tree algorithms)
1,二叉樹(Binary tree)
二叉樹:每一個節點最多兩個子節點,如下圖所示:
相關概念:節點Node,路徑path,根節點root,邊edge,子節點 children,父節點parent,兄弟節點sibling, 子樹subtree,葉子節點leaf node, 度level,樹高hight
節點Node: 路徑path:從一個節點到擰一個節點間的邊 根節點root, 邊edge:節點間的連線 子節點 children, 父節點parent, 兄弟節點sibling, 子樹subtree, 葉子節點leaf node, 度level:從當前節點到根節點的路徑中邊的數量 高度 hight:樹中所有節點的最大levelView Code
二叉樹可以通過多級列表的形式實現,多級列表形式如下,根節點r,有兩個子節點a , b,且a, b節點沒有子節點。
mytree =[ r,
[ a, [ ], [ ] ], [ b, [ ], [ ] ]
]
python實現程式碼如下:
#coding:utf-8 #多級列表實現 def binaryTree(r): return [r,[],[]] #root[]為根節點,root[1]左子樹,root[2]右子樹多級列表形式def insertLeftTree(root,newbranch): t = root.pop(1) if len(t)>1: root.insert(1, [newbranch, t, []]) else: root.insert(1,[newbranch, [], []]) return root def insertRightTree(root,newbranch): t = root.pop(2) if len(t)>1: root.insert(2, [newbranch, [], t])else: root.insert(2,[newbranch, [], []]) return root def getRootVal(root): return root[0] def setRootVal(root,val): root[0]= val def getLeftChildren(root): return root[1] def getRightChildren(root): return root[2] r = binaryTree(3) insertLeftTree(r,4) insertLeftTree(r,5) insertRightTree(r,6) insertRightTree(r,7) l = getLeftChildren(r) print(l) setRootVal(l,9) print(r) insertLeftTree(l,11) print(r) print(getRightChildren(getRightChildren(r)))
二叉樹可以通過節點的形式實現,如下所示:
python實現程式碼如下:
class BinaryTree(object): def __init__(self,value): self.key = value self.leftChild = None self.rightChild = None def insertLeft(self,newNode): if self.leftChild != None: temp = BinaryTree(newNode) temp.leftChild = self.leftChild self.leftChild = temp else: self.leftChild = BinaryTree(newNode) def insertRight(self,newNode): if self.rightChild != None: temp = BinaryTree(newNode) temp.rightChild= self.rightChild self.rightChild = temp else: self.rightChild = BinaryTree(newNode) def getRootVal(self): return self.key def setRootVal(self,value): self.key = value def getLeftChild(self): return self.leftChild def getRightChild(self): return self.rightChild節點形式
2,二叉樹的應用
2.1 解析樹(parse tree)
解析樹常用於表示真實世界的結構表示,如句子和數學表示式。如下圖是((7+3)*(5-2))的解析樹表示,根據解析樹的層級結構,從下往上計算,能很好的代替括號的表示式中括號的作用
將一個全括號數學表示式轉化為解析樹的過程如下:
遍歷表示式:
1,若碰到“(”,為當前節點插入左節點,並移動到左節點
2,若碰到 + ,- ,* , /,設定當前節點的值為該符號,併為當前節點插入右節點,並移動到右節點
3,若碰到數字,設定當前節點的值為該數字,並移動到其父節點
4,若碰到“)”,移動到當前節點的父節點
python實現程式碼如下:(Stack 參見資料結構之棧 )
from stackDemo import Stack #參見資料結構之棧 def buildParseTree(expstr): explist = expstr.split() s = Stack() t = BinaryTree('') s.push(t) current = t for token in explist: #token = token.strip() if token =='(': current.insertLeft('') s.push(current) current = current.getLeftChild() elif token in ['*','/','+','-']: current.setRootVal(token) current.insertRight('') s.push(current) current = current.getRightChild() elif token not in ['(','*','/','+','-',')']: current.setRootVal(token) current = s.pop() elif token==')': current = s.pop() else: raise ValueError return t t = buildParseTree("( ( 10 + 5 ) * 3 )")構造解析樹
計算解析樹:數學表示式轉化為解析樹後,可以對其進行計算,python程式碼如下:
import operator def evaluate(parseTree): operators={'+':operator.add,'-':operator.sub,'*':operator.mul,'/':operator.div } rootval = parseTree.getRootVal() left = parseTree.getLeftChild() right = parseTree.getRightChild() if left and right: fn = operators[rootval] return fn(evaluate(left),evaluate(right)) else: return parseTree.getRootVal()計算解析樹
中序遍歷解析樹,可以將其還原為全括號數學表示式,python程式碼如下:
#解析樹轉換為全括號數學表示式 def printexp(tree): val = '' if tree: val = '('+printexp(tree.getLeftChild()) val = val +str(tree.getRootVal()) val = val +printexp(tree.getRightChild())+')' if tree.getLeftChild()==None and tree.getRightChild()==None: val = val.strip('()') return val t = buildParseTree("( ( 10 + 5 ) * 3 )") exp = printexp(t) print expView Code
3,樹的遍歷
樹的遍歷包括前序遍歷(preorder),中序遍歷(inorder)和後序遍歷(postorder).
前序遍歷:先訪問根節點,再訪問左子樹,最後訪問右子樹(遞迴),python程式碼實現如下:
def preorder(tree): if tree: print tree.getRootVal() preorder(tree.getLeftChild()) preorder(tree.getRightChild()) #定義在類中的前序遍歷 # def preorder(self): # print self.key # if self.leftChild: # self.leftChild.preorder() # if self.rightChild: # self.rightChild.preorder()preorder
中序遍歷:先訪問左子樹,再訪問根節點,最後訪問右子樹(遞迴),python程式碼實現如下:
#中序遍歷inorder def inorder(tree): if tree: preorder(tree.getLeftChild()) print tree.getRootVal() preorder(tree.getRightChild())View Code
後續遍歷:先訪問左子樹,再訪問右子樹,最後訪問根節點,python程式碼實現如下:
def postorder(tree): if tree : postorder(tree.getLeftChild()) postorder(tree.getRightChild()) print(tree.getRootVal())View Code
4,優先佇列和二叉樹(priority queue and binary heap)
優先佇列:優先佇列和佇列類似,enqueue操作能加入元素到佇列末尾,dequeue操作能移除佇列首位元素,不同的是優先佇列的元素具有優先順序,首位元素具有最高或最小優先順序,因此當進行enqueue操作時,還需要根據元素的優先順序將其移動到適合的位置。優先佇列一般利用二叉堆來實現,其enqueue和dequeue的複雜度都為O(logn)。(也可以用list來實現,但list的插入複雜度為O(n),再進行排序的複雜度為O(n logn))
二叉堆:二叉堆是一顆完全二叉樹,當父節點的鍵值總是大於或等於任何一個子節點的鍵值時為最大堆,當父節點的鍵值總是小於或等於任何一個子節點的鍵值時為最小堆。(完全二叉樹:除最後一層外,每一層上的節點數均達到最大值;在最後一層上只缺少右邊的若干結點;滿二叉樹:除葉子結點外的所有結點均有兩個子結點。節點數達到最大值。所有葉子結點必須在同一層上)
最小堆示例及操作如下:(父節點的值總是小於或等於子節點)
BinaryHeap() #建立空的二叉堆 insert(k) #插入新元素 findMin() #返回最小值,不刪除 delMin() #返回最小值,並刪除 isEmpty() size() buildHeap(list) #通過list建立二叉堆View Code
對於完全二叉樹,若根節點的序號為p,則左右節點的序號應該為2p和2p+1,結合上圖可以發現,可以用一個佇列(首位元素為0)來表示二叉堆的結構。最小堆的python實現程式碼如下:(heaplist中第一個元素為0,不會用到,只是為了保證二叉堆的序列從1開始,方便進行除和乘2p,2p+1)
#coding:utf-8 class BinaryHeap(object): def __init__(self): self.heapList=[0] self.size = 0 #將元素加到完全二叉樹末尾,然後再根據其大小調整其位置 def insert(self,k): self.heapList.append(k) self.size = self.size+1 self._percUp(self.size) # 如果當前節點比父節點小,和父節點交換位置,一直向上重複該過程 def _percUp(self,size): i = size while i>0: if self.heapList[i]<self.heapList[i//2]: temp = self.heapList[i] self.heapList[i] = self.heapList[i//2] self.heapList[i//2] = temp i=i//2 # 將根元素返回,並將最末尾元素移動到根元素保持完全二叉樹結構不變,再根據大小,將新的根元素向下移動到合適的位置 def delMin(self): temp = self.heapList[1] self.heapList[1]=self.heapList[self.size] self.size = self.size-1 self.heapList.pop() self._percDown(1) return temp # 如果當前節點比最小子節點大,和該子節點交換位置,一直向下重複該過程 def _percDown(self,i): while (2*i)<=self.size: mc = self._minChild(i) if self.heapList[i]>self.heapList[mc]: temp = self.heapList[i] self.heapList[i]=self.heapList[mc] self.heapList[mc] =temp i = mc #返回左右子節點中較小子節點的位置 def _minChild(self,i): if (2*i+1)>self.size: return 2*i else: if self.heapList[2*i] < self.heapList[2*i+1]: return 2*i else: return 2*i+1 #通過一個list建立二叉堆 def buildHeap(self,list): i = len(list)//2 self.heapList = [0]+list[:] self.size = len(list) while i>0: self._percDown(i) i = i-1View Code
insert()插入過程示例圖如下:將元素加到完全二叉樹末尾,然後再根據其大小調整其位置
delMin()操作過程示例如下:將根元素返回,並將最末尾元素移動到根元素保持完全二叉樹結構不變,再根據大小,將新的根元素向下移動到合適的位置
insert和delMin的複雜度都為O(log n), buildHeap的複雜度為O(n),利用二叉堆對list進行排序,複雜度為O(n log n),程式碼如下:
#通過list構造二叉堆,然後不斷將堆頂元素返回,就得到排序好的list alist = [54,26,93,17,98,77,31,44,55,20] h = BinaryHeap() h.buildHeap(alist) s=[] while h.size>0: s.append(h.delMin()) print sView Code
5,二叉搜尋樹(Binary Search Tree, bst)
二叉搜尋樹:左節點的值,總是小於其父節點的值,右節點的值總是大於其父節點的值(bst property)。如下圖所示:
利用二叉搜尋樹實現map(字典),常用操作如下:
Map() # 建立字典 put(key,val) # 字典中插入資料 get(key) # 取鍵值 del # 刪除 len() # 求長度 in # 是否存在View Code
python實現map程式碼如下:
#coding:utf-8 class TreeNode(object): def __init__(self,key, value, leftChild=None,rightChild=None,parent=None): self.key = key self.value = value self.leftChild = leftChild self.rightChild = rightChild self.parent = parent self.balanceFactor =0 def hasLeftChild(self): return self.leftChild def hasRightChild(self): return self.rightChild def isLeftChild(self): return self.parent and self.parent.leftChild==self def isRightChild(self): return self.parent and self.parent.rightChild==self def isRoot(self): return not self.parent def isLeaf(self): return not (self.leftChild or self.rightChild) def hasAnyChildren(self): return self.leftChild or self.rightChild def hasBothChildren(self): return self.leftChild and self.rightChild def replaceNodeData(self,key,value,lc=None,rc=None): self.key=key self.value = value self.leftChild = lc self.rightChild = rc if self.hasLeftChild(): self.leftChild.parent = self if self.hasRightChild(): self.rightChild = self def __iter__(self): if self: if self.hasLeftChild(): for elem in self.leftChild: #呼叫self.leftChiLd.__iter__(),所以此處是遞迴的 yield elem yield self.key, self.value, self.balanceFactor if self.hasRightChild(): for elem in self.rightChild: #呼叫self.rightChiLd.__iter__() yield elem def findSuccessor(self): #尋找繼承 succ = None if self.hasRightChild(): succ = self.rightChild._findMin() else: if self.parent: if self.isLeftChild(): succ = self.parent else: self.parent.rightChild = None succ = self.parent.findSuccessor() self.parent.rightChild = self return succ def _findMin(self): current = self while current.hasLeftChild(): current = current.leftChild return current def spliceOut(self): if self.isLeaf(): if self.isLeftChild(): self.parent.leftChild=None else: self.parent.rightChild=None elif self.hasAnyChildren(): if self.hasLeftChild(): if self.isLeftChild(): self.parent.leftChild = self.leftChild else: self.parent.rightChild = self.leftChild self.leftChild.parent = self.parent else: if self.isLeftChild(): self.parent.leftChild = self.rightChild else: self.parent.rightChild = self.rightChild self.rightChild.parent = self.parent class BinarySearchTree(object): def __init__(self): self.root = None self.size = 0 def length(self): return self.size def __len__(self): return self.size def __iter__(self): return self.root.__iter__() #加入元素 def put(self,key,value): if self.root: self._put(key,value,self.root) else: self.root = TreeNode(key,value) self.size = self.size+1 def _put(self,key,value,currentNode): if currentNode.key<key: if currentNode.hasRightChild(): self._put(key,value,currentNode.rightChild) else: currentNode.rightChild=TreeNode(key,value,parent=currentNode) elif currentNode.key>key: if currentNode.hasLeftChild(): self._put(key,value,currentNode.leftChild) else: currentNode.leftChild=TreeNode(key,value,parent=currentNode) else: currentNode.replaceNodeData(key,value) def __setitem__(self, key, value): self.put(key,value) #獲取元素值 def get(self,key): if self.root: node = self._get(key,self.root) if node: return node.value else: return None else: return None def _get(self,key,currentNode): if not currentNode: return None if currentNode.key==key: return currentNode elif currentNode.key<key: return self._get(key,currentNode.rightChild) #rightChild可能不存在 else: return self._get(key,currentNode.leftChild) #leftChild可能不存在 # def _get(self,key,currentNode): # if currentNode.key == key: # return currentNode # elif currentNode.key<key: # if currentNode.hasRightChild(): # return self._get(key,currentNode.rightChild) # else: # return None # else: # if currentNode.hasLeftChild(): # return self._get(key,currentNode.leftChild) # else: # return None def __getitem__(self, key): return self.get(key) def __contains__(self, key): #實現 in 操作 if self._get(key,self.root): return True else: return False def delete(self,key): if self.size>1: node = self._get(key,self.root) if node: self._del(node) self.size = self.size - 1 else: raise KeyError('Error, key not in tree') elif self.size==1 and self.root.key==key: self.root = None self.size = self.size - 1 else: raise KeyError('Error, key not in tree') def _del(self,currentNode): if currentNode.isLeaf(): if currentNode.isLeftChild(): currentNode.parent.leftChild = None elif currentNode.isRightChild(): currentNode.parent.rightChild = None elif currentNode.hasBothChildren(): successor = currentNode.findSuccessor() #此處successor為其右子樹的最小值,即最左邊的值 successor.spliceOut() currentNode.key = successor.key currentNode.value = successor.value elif currentNode.hasAnyChildren(): if currentNode.hasLeftChild(): if currentNode.isLeftChild(): currentNode.parent.leftChild = currentNode.leftChild currentNode.leftChild.parent = currentNode.parent elif currentNode.isRightChild(): currentNode.parent.rightChild = currentNode.leftChild currentNode.leftChild.parent = currentNode.parent else: # currentNode has no parent (is root) currentNode.replaceNodeData(currentNode.leftChild.key, currentNode.leftChild.value, currentNode.leftChild.leftChild, currentNode.leftChild.rightChild) elif currentNode.hasRightChild(): if currentNode.isLeftChild(): currentNode.parent.leftChild = currentNode.rightChild currentNode.rightChild.parent = currentNode.parent elif currentNode.isRightChild(): currentNode.parent.rightChild = currentNode.rightChild currentNode.rightChild.parent = currentNode.parent else: # currentNode has no parent (is root) currentNode.replaceNodeData(currentNode.rightChild.key, currentNode.rightChild.value, currentNode.rightChild.leftChild, currentNode.rightChild.rightChild) def __delitem__(self, key): self.delete(key) if __name__ == '__main__': mytree = BinarySearchTree() mytree[8]="red" mytree[4]="blue" mytree[6]="yellow" mytree[5]="at" mytree[9]="cat" mytree[11]="mat" print(mytree[6]) print(mytree[5]) for x in mytree: print x del mytree[6] print '-'*12 for x in mytree: print xView Code
在上述程式碼中最複雜的為刪除操作,刪除節點時有三種情況:節點為葉子節點,節點有兩個子節點,節點有一個子節點。當節點有兩個子節點時,對其刪除時,應該用其右子樹的最小值來代替其位置(即右子樹中最左邊的值)。
對於map進行復雜度分析,可以發現put,get取決於tree的高度,當節點隨機分配時複雜度為O(log n),但當節點分佈不平衡時,複雜度會變成O(n),如下圖所示:
6, 平衡二叉搜尋樹 (Balanced binary search tree, AVL tree)
平衡二叉搜尋樹:又稱為AVL Tree,取名於發明者G.M. Adelson-Velskii 和E.M. Landis,在二叉搜尋樹的基礎上引入平衡因子(balance factor),每次插入和刪除節點時都保持樹平衡,從而避免上面出現的搜尋二叉樹複雜度會變成O(n)。一個節點的balance factor的計算公式如下,即該節點的左子樹高度減去右子樹高度。
當樹所有節點的平衡因子為-1,0,1時,該樹為平衡樹,平衡因子大於1或小於-1時,樹不平衡需要調整,下圖為一顆樹的各個節點的平衡因子。(1時樹left-heavy,0時完全平衡,-1時right-heavy)
相比於二叉搜尋樹,AVL樹的put和delete操作後,需要對節點的平衡因子進行更新,如果某個節點不平衡時,需要進行平衡處理,主要分為左旋轉和右旋轉。
左旋轉:如圖,節點A的平衡因子為-2(right heavy),不平衡,對其進行左旋轉,即以A為旋轉點,AB邊逆時針旋轉。
詳細操作為:1,A的右節點B作為新的子樹根節點
2,A成為B的左節點,如果B有左節點時,將其左節點變為A的右節點(A的右節點原來為B,所以A的右節點現在為空)
右旋轉:如圖,節點E的平衡因子為2(left heavy),不平衡,對其進行右旋轉,即以E為旋轉點,EC邊順時針旋轉。
詳細操作為:1,E的左節點C作為新的子樹根節點
2,E成為C的右節點,如果C有右節點時,將其右節點變為E的左節點(E的左節點原來為C,所以E的左節點現在為空)
特殊情況:當出現下面的情況時,如圖所示,A依舊為right heavy,但若進行左旋轉,又會出現left heavy,無法完成平衡操作。 所以在進行左旋轉和右旋轉前需要進行一步判斷,具體操作如下:
1,如果某節點需要進行左旋轉平衡時(right heavy),檢查其右子節點的平衡因子,若右子節點為left heavy,先對右子節點右旋轉,然後對該節點左旋轉
2,如果某節點需要進行右旋轉平衡時(left heavy),檢查其左子節點的平衡因子,若左子節點為right heavy,先對左子節點左旋轉,然後對該節點右旋轉
AVL tree用python實現的程式碼如下:
#coding:utf-8 from binarySearchTree import TreeNode, BinarySearchTree # class AVLTreeNode(TreeNode): # # def __init__(self,*args,**kwargs): # self.balanceFactor = 0 # super(AVLTreeNode,self).__init__(*args,**kwargs) class AVLTree(BinarySearchTree): def _put(self,key,value,currentNode): if currentNode.key<key: if currentNode.hasRightChild(): self._put(key,value,currentNode.rightChild) else: currentNode.rightChild=TreeNode(key,value,parent=currentNode) self.updateBalance(currentNode.rightChild) elif currentNode.key>key: if currentNode.hasLeftChild(): self._put(key,value,currentNode.leftChild) else: currentNode.leftChild=TreeNode(key,value,parent=currentNode) self.updateBalance(currentNode.leftChild) else: currentNode.replaceNodeData(key,value) def _del(self,currentNode): if currentNode.isLeaf(): if currentNode.isLeftChild(): currentNode.parent.leftChild = None currentNode.parent.balanceFactor -=1 elif currentNode.isRightChild(): currentNode.parent.rightChild = None currentNode.parent.balanceFactor += 1 if currentNode.parent.balanceFactor>1 or currentNode.parent.balanceFactor<-1: self.reblance(currentNode.parent) elif currentNode.hasBothChildren(): successor = currentNode.findSuccessor() #此處successor為其右子樹的最小值,即最左邊的值 # 先更新parent的balanceFactor if successor.isLeftChild(): successor.parent.balanceFactor -= 1 elif successor.isRightChild(): successor.parent.balanceFactor += 1 successor.spliceOut() currentNode.key = successor.key currentNode.value = successor.value # 刪除後,再判斷是否需要再平衡,然後進行再平衡操作 if successor.parent.balanceFactor>1 or successor.parent.balanceFactor<-1: self.reblance(successor.parent) elif currentNode.hasAnyChildren(): #先更新parent的balanceFactor if currentNode.isLeftChild(): currentNode.parent.balanceFactor -= 1 elif currentNode.isRightChild(): currentNode.parent.balanceFactor += 1 if currentNode.hasLeftChild(): if currentNode.isLeftChild(): currentNode.parent.leftChild = currentNode.leftChild currentNode.leftChild.parent = currentNode.parent elif currentNode.isRightChild(): currentNode.parent.rightChild = currentNode.leftChild currentNode.leftChild.parent = currentNode.parent else: # currentNode has no parent (is root) currentNode.replaceNodeData(currentNode.leftChild.key, currentNode.leftChild.value, currentNode.leftChild.leftChild, currentNode.leftChild.rightChild) elif currentNode.hasRightChild(): if currentNode.isLeftChild(): currentNode.parent.leftChild = currentNode.rightChild currentNode.rightChild.parent = currentNode.parent elif currentNode.isRightChild(): currentNode.parent.rightChild = currentNode.rightChild currentNode.rightChild.parent = currentNode.parent else: # currentNode has no parent (is root) currentNode.replaceNodeData(currentNode.rightChild.key, currentNode.rightChild.value, currentNode.rightChild.leftChild, currentNode.rightChild.rightChild) #刪除後,再判斷是否需要再平衡,然後進行再平衡操作 if currentNode.parent!=None: #不是根節點 if currentNode.parent.balanceFactor>1 or currentNode.parent.balanceFactor<-1: self.reblance(currentNode.parent) def updateBalance(self,node): if node.balanceFactor>1 or node.balanceFactor<-1: self.reblance(node) return if node.parent!=None: if node.isLeftChild(): node.parent.balanceFactor +=1 elif node.isRightChild(): node.parent.balanceFactor -=1 if node.parent.balanceFactor!=0: self.updateBalance(node.parent) def reblance(self,node): if node.balanceFactor>1: if node.leftChild.balanceFactor<0: self.rotateLeft(node.leftChild) self.rotateRight(node) elif node.balanceFactor<-1: if node.rightChild.balanceFactor>0: self.rotateRight(node.rightChild) self.rotateLeft(node) def rotateLeft(self,node): newroot = node.rightChild node.rightChild = newroot.leftChild if newroot.hasLeftChild(): newroot.leftChild.parent = node newroot.parent = node.parent if node.parent!=None: if node.isLeftChild(): node.parent.leftChild = newroot elif node.isRightChild(): node.parent.rightChild = newroot else: self.root = newroot newroot.leftChild = node node.parent = newroot node.balanceFactor = node.balanceFactor+1-min(newroot.balanceFactor,0) newroot.balanceFactor = newroot.balanceFactor+1+max(node.balanceFactor,0) def rotateRight(self,node): newroot = node.leftChild node.leftChild = newroot.rightChild if newroot.rightChild!=None: newroot.rightChild.parent = node newroot.parent = node.parent if node.parent!=None: if node.isLeftChild(): node.parent.leftChild = newroot elif node.isRightChild(): node.parent.rightChild = newroot else: self.root = newroot newroot.rightChild = node node.parent = newroot node.balanceFactor = node.balanceFactor-1-max(newroot.balanceFactor,0) newroot.balanceFactor = newroot.balanceFactor-1+min(node.balanceFactor,0) if __name__ == '__main__': mytree = AVLTree() mytree[8]="red" mytree[4]="blue" mytree[6]="yellow" mytree[5]="at" mytree[9]="cat" mytree[11]="mat" print(mytree[6]) print(mytree[5]) print '-'*12 print ('key','value','balanceFactor') for x in mytree: print x print 'root:',mytree.root.key del mytree[6] print '-'*12 print ('key','value','balanceFactor') for x in mytree: print x print 'root:',mytree.root.keyView Code
AVL Tree繼承了二叉搜尋樹,對其插入和刪除方法進行了重寫,另外對TreeNode增加了balanceFactor屬性。再進行左旋轉和右旋轉時,對於balanceFactor的需要計算一下,如圖的左旋轉過程中,D成為了新的根節點,只有B和D的平衡因子發生了變化,需要對其進行更新。(右旋轉和左旋轉類似)
B的平衡因子計算過程如下:(newBal(B)為左旋轉後B的平衡因子,oldBal(B)為原來的節點B的平衡因子,h為節點的高度)
D的平衡因子計算過程如下:
由於AVL Tree總是保持平衡,其put和get操作的複雜度能保持為O(log n)
7.總結
到目前為止,對於map(字典)資料結構,用二叉搜尋樹和AVL樹實現了,也用有序列表和雜湊表實現過,對應操作的複雜度如下:
參考:http://interactivepython.org/runestone/static/pythonds/Trees/toctree.html