本文實例講述了Python實現基于二叉樹存儲結構的堆排序算法。分享給大家供大家參考，具體如下：
既然用Python實現了二叉樹，當然要寫點東西練練手。
網絡上堆排序的教程很多，但是卻幾乎都是以數組存儲的數，直接以下標訪問元素，當然這樣是完全沒有問題的，實現簡單，訪問速度快，也容易理解。
但是以練手的角度來看，我還是寫了一個二叉樹存儲結構的堆排序
其中最難的問題就是交換二叉樹中兩個節點。
因為一個節點最多與三個節點相連，那么兩個節點互換，就需要考慮到5個節點之間的關系，也需要判斷是左右孩子，這將是十分繁瑣的，也很容易出錯。
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class Tree: def __init__(self, val = '#', left = None, right = None): self.val = val self.left = left self.right = right self.ponit = None self.father = None self.counter = 0 #前序構建二叉樹 def FrontBuildTree(self): temp = input('Please Input: ') node = Tree(temp) if(temp != '#'): node.left = self.FrontBuildTree() node.right = self.FrontBuildTree() return node#因為沒有引用也沒有指針，所以就把新的節點給返回回去 #前序遍歷二叉樹 def VisitNode(self): print(self.val) if(self.left != None): self.left.VisitNode() if(self.right != None): self.right.VisitNode() #中序遍歷二叉樹 def MVisitTree(self): if(self.left != None): self.left.MVisitTree() print(self.val) if(self.right != None): self.right.MVisitTree() #獲取二叉樹的第dec個節點 def GetPoint(self, dec): road = str(bin(dec))[3:] p = self for r in road: if (r == '0'): p = p.left else: p = p.right #print('p.val = ', p.val) return p #構建第一個堆 def BuildHeadTree(self, List): for val in List: #print('val = ', val, 'self.counter = ', self.counter) self.ponit = self.GetPoint(int((self.counter + 1) / 2)) #print('self.ponit.val = ', self.ponit.val) if (self.counter == 0): self.val = val self.father = self else: temp = self.counter + 1 node = Tree(val) node.father = self.ponit if(temp % 2 == 0):#新增節點為左孩子 self.ponit.left = node else: self.ponit.right = node while(temp != 0): if (node.val < node.father.val):#如果新增節點比其父親節點值要大 p = node.father#先將其三個鏈子保存起來 LeftTemp = node.left RightTemp = node.right if (p.father != p):#判斷其不是頭結點 if (int(temp / 2) % 2 == 0):#新增節點的父親為左孩子 p.father.left = node else: p.father.right = node node.father = p.father else: node.father = node#是頭結點則將其father連向自身 node.counter = self.counter self = node if(temp % 2 == 0):#新增節點為左孩子 node.left = p node.right = p.right if (p.right != None): p.right.father = node else: node.left = p.left node.right = p if (p.left != None): p.left.father = node p.left = LeftTemp p.right = RightTemp p.father = node temp = int(temp / 2) #print('node.val = ', node.val, 'node.father.val = ', node.father.val) #print('Tree = ') #self.VisitNode() else: break; self.counter += 1 return self #將頭結點取出后重新調整堆 def Adjust(self): #print('FrontSelfTree = ') #self.VisitNode() #print('MSelfTree = ') #self.MVisitTree() print('Get ', self.val) p = self.GetPoint(self.counter) #print('p.val = ', p.val) #print('p.father.val = ', p.father.val) root = p if (self.counter % 2 == 0): p.father.left = None else: p.father.right = None #print('self.left = ', self.left.val) #print('self.right = ', self.right.val) p.father = p#將二叉樹最后一個葉子節點移到頭結點 p.left = self.left p.right = self.right while(1):#優化是萬惡之源 LeftTemp = p.left RightTemp = p.right FatherTemp = p.father if (p.left != None and p.right !=None):#判斷此時正在處理的結點的左后孩子情況 if (p.left.val < p.right.val): next = p.left else: next = p.right if (p.val < next.val): break; elif (p.left == None and p.right != None and p.val > p.right.val): next = p.right elif (p.right == None and p.left != None and p.val > p.left.val): next = p.left else: break; p.left = next.left p.right = next.right p.father = next if (next.left != None):#之后就是一系列的交換節點的鏈的處理 next.left.father = p if (next.right != None): next.right.father = p if (FatherTemp == p): next.father = next root = next else: next.father == FatherTemp if (FatherTemp.left == p): FatherTemp.left = next else: FatherTemp.right = next if (next == LeftTemp): next.right = RightTemp next.left = p if (RightTemp != None): RightTemp.father = next else: next.left = LeftTemp next.right = p if (LeftTemp != None): LeftTemp.father = next #print('Tree = ') #root.VisitNode() root.counter = self.counter - 1 return rootif __name__ == '__main__': print("服務器之家測試結果") root = Tree() number = [-1, -1, 0, 0, 0, 12, 22, 3, 5, 4, 3, 1, 6, 9] root = root.BuildHeadTree(number) while(root.counter != 0): root = root.Adjust() |
運行結果：
希望本文所述對大家Python程序設計有所幫助。
原文鏈接：http://blog.csdn.net/yk_ee/article/details/64121486
