如何用python构造一个n层的完全二叉树
2024-05-16 03:39
1. 如何用python构造一个n层的完全二叉树
用python构造一个n层的完全二叉树的代码如下： typedef struct {int weight;int parent, lchild, rchild; } HTNode ,*HuffmanTree; // 动态分配数组存储huffman树 算法设计void createHuffmantree(){ ht=(HuffmanTree)malloc(m+1)*sizeof(HTNode);// 动态分配数组存储huffman树,0号单元未用// m:huffman 树中的结点数（m=2*n-1）for (i=1;ilch= ht[i]->rch=0; for (i=1;i<=n;++i) ht[i].weight=w[i]; //初始化，w[i]：n个叶子的权值 for (i=n+1;i<=m,++i) ｛ //建哈夫曼树 select(i-1),s1,s2); //在ht[k](1<=k<=i-1)中选择两个双亲域为零而权值取最小的结点 :s1和s2 ht[s1].parent= ht[s2].parent=i; ht[i].lch=s1; ht[i].rch=s2; ht[i].weight=ht[s1].weight + ht[s2].weight ; };}
2. python 二叉树是怎么实现的
#coding:utf-8#author:Elvis class TreeNode(object): def __init__(self): self.data = '#' self.l_child = None self.r_child = None class Tree(TreeNode): #create a tree def create_tree(self, tree): data = raw_input('->') if data == '#': tree = None else: tree.data = data tree.l_child = TreeNode() self.create_tree(tree.l_child) tree.r_child = TreeNode() self.create_tree(tree.r_child) #visit a tree node def visit(self, tree): #输入#号代表空树 if tree.data is not '#': print str(tree.data) + '\t', #先序遍历 def pre_order(self, tree): if tree is not None: self.visit(tree) self.pre_order(tree.l_child) self.pre_order(tree.r_child) #中序遍历 def in_order(self, tree): if tree is not None: self.in_order(tree.l_child) self.visit(tree) self.in_order(tree.r_child) #后序遍历 def post_order(self, tree): if tree is not None: self.post_order(tree.l_child) self.post_order(tree.r_child) self.visit(tree) t = TreeNode()tree = Tree()tree.create_tree(t)tree.pre_order(t)print '\n'tree.in_order(t)print '\n'tree.post_order(t)
3. python二叉树算法
定义一颗二叉树，请看官自行想象其形状
class BinNode( ):
def __init__( self, val ):
self.lchild = None
self.rchild = None
self.value = val
binNode1 = BinNode( 1 )
binNode2 = BinNode( 2 )
binNode3 = BinNode( 3 )
binNode4 = BinNode( 4 )
binNode5 = BinNode( 5 )
binNode6 = BinNode( 6 )
binNode1.lchild = binNode2
binNode1.rchild = binNode3
binNode2.lchild = binNode4
binNode2.rchild = binNode5
binNode3.lchild = binNode6
4. python字典怎么表现二叉树
用python构造一个n层的完全二叉树的代码如下： typedef struct {int weight;int parent, lchild, rchild; } HTNode ,*HuffmanTree; // 动态分配数组存储huffman树 算法设计void createHuffmantree(){ ht=(HuffmanTree)malloc(m+1)*sizeof(HTNode.
5. 如何将数据存储为二叉树python
（1）二叉树是有序树，即使只有一个子树，也必须区分左、右子树； （2）二叉树的每个结点的度不能大于2，只能取0、1、2三者之一； （3）二叉树中所有结点的形态有5种：空结点、无左右子树的结点、只有左子树的结点、只有右子树的结点和具有左右子树的结点。
6. python怎么获得二叉树根到所有叶子的路径
一棵树当中没有子结点（即度为0）的结点，称为叶子结点因为只有一个所以1 2 3 4是一直这样连下去 的
7. python 查找二叉树是否有子树
python中的二叉树模块内容： BinaryTree：非平衡二叉树 AVLTree：平衡的AVL树 RBTree：平衡的红黑树 以上是用python写的，相面的模块是用c写的，并且可以做为Cython的包。 FastBinaryTree FastAVLTree FastRBTree 特别需要说明的是：树往往要比python内置的dict类慢一些，但是它中的所有数据都是按照某个关键词进行排序的，故在某些情况下是必须使用的。 安装和使用 安装方法 安装环境： ubuntu12.04, python 2.7.6
8. 求Python二叉树的几个算法 求几个二叉树的method！ 1) 给一个值，然后在树中找出该值
你好：
二叉树算法，网上是比较多的；
可能按照你的需求不是很多：
下面是我用的一个，不过你可以借鉴一下的：
# -*- coding: cp936 -*-import osclass Node(object): """docstring for Node""" def __init__(self, v = None, left = None, right=None, parent=None): self.value = v self.left = left self.right = right self.parent = parentclass BTree(object): """docstring for BtTee """ def __init__(self): self.root = None self.size = 0 def insert(self, node): n = self.root if n == None: self.root = node return while True: if node.value n.value: if n.right == None: n.parent = n n.right = node break else: n = n.right def find(self, v): n = self.root # http://yige.org while True: if n == None: return None if v == n.value: return n if v n.value: n = n.right def find_successor(node): '''查找后继结点''' assert node != None and node.right != None n = node.right while n.left != None: n = n.left return n def delete(self, v): n = self.find(v) print "delete:",n.value del_parent = n.parent if del_parent == None: self.root = None; return if n != None: if n.left != None and n.right != None: succ_node = find_successor(n) parent = succ_node.parent if succ_node == parent.left: #if succ_node is left sub tree parent.left = None if succ_node == parent.right: #if succ_node is right sub tree parent.right = None if del_parent.left == n: del_parent.left = succ_node if del_parent.right == n: del_parent.right = succ_node succ_node.parent = n.parent succ_node.left = n.left succ_node.right = n.right del n elif n.left != None or n.right != None: if n.left != None: node = n.left else: node = n.right node.parent = n.parent if del_parent.left == n: del_parent.left = node if del_parent.right == n: del_parent.right = node del n else: if del_parent.left == n: del_parent.left = None if del_parent.right == n: del_parent.right = None def tranverse(self): def pnode(node): if node == None: return if node.left != None: pnode(node.left) print node.value if node.right != None: pnode(node.right) pnode(self.root)def getopts(): import optparse, locale parser = optparse.OptionParser() parser.add_option("-i", "--input", dest="input", help=u"help name", metavar="INPUT") (options, args) = parser.parse_args() #print options.input return (options.input) if __name__ == '__main__': al = [23, 45, 67, 12, 78,90, 11, 33, 55, 66, 89, 88 ,5,6,7,8,9,0,1,2,678] bt = BTree() for x in al : bt.insert(Node(x)) bt.delete(12) bt.tranverse() n = bt.find(12) if n != None: print "find valud:",n.value
最新文章
热门文章
推荐阅读