Pythonでバイナリツリーを実装するために使用できる最適なデータ構造はどれですか?
ここに、バイナリツリーの単純な再帰的な実装を示します。
#!/usr/bin/python
class Node:
def __init__(self, val):
self.l = None
self.r = None
self.v = val
class Tree:
def __init__(self):
self.root = None
def getRoot(self):
return self.root
def add(self, val):
if(self.root == None):
self.root = Node(val)
else:
self._add(val, self.root)
def _add(self, val, node):
if(val < node.v):
if(node.l != None):
self._add(val, node.l)
else:
node.l = Node(val)
else:
if(node.r != None):
self._add(val, node.r)
else:
node.r = Node(val)
def find(self, val):
if(self.root != None):
return self._find(val, self.root)
else:
return None
def _find(self, val, node):
if(val == node.v):
return node
Elif(val < node.v and node.l != None):
self._find(val, node.l)
Elif(val > node.v and node.r != None):
self._find(val, node.r)
def deleteTree(self):
# garbage collector will do this for us.
self.root = None
def printTree(self):
if(self.root != None):
self._printTree(self.root)
def _printTree(self, node):
if(node != None):
self._printTree(node.l)
print str(node.v) + ' '
self._printTree(node.r)
# 3
# 0 4
# 2 8
tree = Tree()
tree.add(3)
tree.add(4)
tree.add(0)
tree.add(8)
tree.add(2)
tree.printTree()
print (tree.find(3)).v
print tree.find(10)
tree.deleteTree()
tree.printTree()
# simple binary tree
# in this implementation, a node is inserted between an existing node and the root
class BinaryTree():
def __init__(self,rootid):
self.left = None
self.right = None
self.rootid = rootid
def getLeftChild(self):
return self.left
def getRightChild(self):
return self.right
def setNodeValue(self,value):
self.rootid = value
def getNodeValue(self):
return self.rootid
def insertRight(self,newNode):
if self.right == None:
self.right = BinaryTree(newNode)
else:
tree = BinaryTree(newNode)
tree.right = self.right
self.right = tree
def insertLeft(self,newNode):
if self.left == None:
self.left = BinaryTree(newNode)
else:
tree = BinaryTree(newNode)
tree.left = self.left
self.left = tree
def printTree(tree):
if tree != None:
printTree(tree.getLeftChild())
print(tree.getNodeValue())
printTree(tree.getRightChild())
# test tree
def testTree():
myTree = BinaryTree("Maud")
myTree.insertLeft("Bob")
myTree.insertRight("Tony")
myTree.insertRight("Steven")
printTree(myTree)
詳細はこちら:-これは非常に単純な 実装 のバイナリツリーです。
これ は、間に質問がある素敵なチュートリアルです
PythonでのBSTの簡単な実装
class TreeNode:
def __init__(self, value):
self.left = None
self.right = None
self.data = value
class Tree:
def __init__(self):
self.root = None
def addNode(self, node, value):
if(node==None):
self.root = TreeNode(value)
else:
if(value<node.data):
if(node.left==None):
node.left = TreeNode(value)
else:
self.addNode(node.left, value)
else:
if(node.right==None):
node.right = TreeNode(value)
else:
self.addNode(node.right, value)
def printInorder(self, node):
if(node!=None):
self.printInorder(node.left)
print(node.data)
self.printInorder(node.right)
def main():
testTree = Tree()
testTree.addNode(testTree.root, 200)
testTree.addNode(testTree.root, 300)
testTree.addNode(testTree.root, 100)
testTree.addNode(testTree.root, 30)
testTree.printInorder(testTree.root)
リストを使用してバイナリツリーを実装する非常に簡単な方法です。最も効率的ではなく、nil値をうまく処理しません。しかし、それは非常に透明です(少なくとも私にとって):
def _add(node, v):
new = [v, [], []]
if node:
left, right = node[1:]
if not left:
left.extend(new)
Elif not right:
right.extend(new)
else:
_add(left, v)
else:
node.extend(new)
def binary_tree(s):
root = []
for e in s:
_add(root, e)
return root
def traverse(n, order):
if n:
v = n[0]
if order == 'pre':
yield v
for left in traverse(n[1], order):
yield left
if order == 'in':
yield v
for right in traverse(n[2], order):
yield right
if order == 'post':
yield v
反復可能からツリーを構築する:
>>> tree = binary_tree('A B C D E'.split())
>>> print tree
['A', ['B', ['D', [], []], ['E', [], []]], ['C', [], []]]
木を横断する:
>>> list(traverse(tree, 'pre')), list(traverse(tree, 'in')), list(traverse(tree, 'post'))
(['A', 'B', 'D', 'E', 'C'],
['D', 'B', 'E', 'A', 'C'],
['D', 'E', 'B', 'C', 'A'])
ここでの答えのほとんどがバイナリ検索ツリーの実装であることに気づかずにはいられません。バイナリ検索ツリー!=バイナリツリー。
バイナリ検索ツリーには非常に特殊なプロパティがあります。ノードXの場合、Xのキーは左の子の子孫のキーよりも大きく、右の子の子孫のキーよりも小さくなります。
二分木はそのような制限を課しません。バイナリツリーは、単に「キー」要素と、「左」と「右」という2つの子を持つデータ構造です。
ツリーは、各ノードが任意の数の子を持つことができるバイナリツリーのさらに一般的なケースです。通常、各ノードには、リスト/配列タイプの「子」要素があります。
さて、OPの質問に答えるために、Pythonでのバイナリツリーの完全な実装を含めます。各BinaryTreeNodeを格納する基礎となるデータ構造は、最適なO(1)ルックアップを提供する場合、辞書です。また、深さ優先および幅優先のトラバーサルも実装しました。これらは、ツリーで実行される非常に一般的な操作です。
from collections import deque
class BinaryTreeNode:
def __init__(self, key, left=None, right=None):
self.key = key
self.left = left
self.right = right
def __repr__(self):
return "%s l: (%s) r: (%s)" % (self.key, self.left, self.right)
def __eq__(self, other):
if self.key == other.key and \
self.right == other.right and \
self.left == other.left:
return True
else:
return False
class BinaryTree:
def __init__(self, root_key=None):
# maps from BinaryTreeNode key to BinaryTreeNode instance.
# Thus, BinaryTreeNode keys must be unique.
self.nodes = {}
if root_key is not None:
# create a root BinaryTreeNode
self.root = BinaryTreeNode(root_key)
self.nodes[root_key] = self.root
def add(self, key, left_key=None, right_key=None):
if key not in self.nodes:
# BinaryTreeNode with given key does not exist, create it
self.nodes[key] = BinaryTreeNode(key)
# invariant: self.nodes[key] exists
# handle left child
if left_key is None:
self.nodes[key].left = None
else:
if left_key not in self.nodes:
self.nodes[left_key] = BinaryTreeNode(left_key)
# invariant: self.nodes[left_key] exists
self.nodes[key].left = self.nodes[left_key]
# handle right child
if right_key == None:
self.nodes[key].right = None
else:
if right_key not in self.nodes:
self.nodes[right_key] = BinaryTreeNode(right_key)
# invariant: self.nodes[right_key] exists
self.nodes[key].right = self.nodes[right_key]
def remove(self, key):
if key not in self.nodes:
raise ValueError('%s not in tree' % key)
# remove key from the list of nodes
del self.nodes[key]
# if node removed is left/right child, update parent node
for k in self.nodes:
if self.nodes[k].left and self.nodes[k].left.key == key:
self.nodes[k].left = None
if self.nodes[k].right and self.nodes[k].right.key == key:
self.nodes[k].right = None
return True
def _height(self, node):
if node is None:
return 0
else:
return 1 + max(self._height(node.left), self._height(node.right))
def height(self):
return self._height(self.root)
def size(self):
return len(self.nodes)
def __repr__(self):
return str(self.traverse_inorder(self.root))
def bfs(self, node):
if not node or node not in self.nodes:
return
reachable = []
q = deque()
# add starting node to queue
q.append(node)
while len(q):
visit = q.popleft()
# add currently visited BinaryTreeNode to list
reachable.append(visit)
# add left/right children as needed
if visit.left:
q.append(visit.left)
if visit.right:
q.append(visit.right)
return reachable
# visit left child, root, then right child
def traverse_inorder(self, node, reachable=None):
if not node or node.key not in self.nodes:
return
if reachable is None:
reachable = []
self.traverse_inorder(node.left, reachable)
reachable.append(node.key)
self.traverse_inorder(node.right, reachable)
return reachable
# visit left and right children, then root
# root of tree is always last to be visited
def traverse_postorder(self, node, reachable=None):
if not node or node.key not in self.nodes:
return
if reachable is None:
reachable = []
self.traverse_postorder(node.left, reachable)
self.traverse_postorder(node.right, reachable)
reachable.append(node.key)
return reachable
# visit root, left, then right children
# root is always visited first
def traverse_preorder(self, node, reachable=None):
if not node or node.key not in self.nodes:
return
if reachable is None:
reachable = []
reachable.append(node.key)
self.traverse_preorder(node.left, reachable)
self.traverse_preorder(node.right, reachable)
return reachable
あなたは2つのクラスを持つ必要はありません
class Tree:
val = None
left = None
right = None
def __init__(self, val):
self.val = val
def insert(self, val):
if self.val is not None:
if val < self.val:
if self.left is not None:
self.left.insert(val)
else:
self.left = Tree(val)
Elif val > self.val:
if self.right is not None:
self.right.insert(val)
else:
self.right = Tree(val)
else:
return
else:
self.val = val
print("new node added")
def showTree(self):
if self.left is not None:
self.left.showTree()
print(self.val, end = ' ')
if self.right is not None:
self.right.showTree()
[インタビューに必要なもの] Nodeクラスは、バイナリツリーを表すのに最低限必要なものです。
(他の答えはほとんど正解ですが、バイナリツリーではオブジェクトクラスを拡張する必要も、BSTである必要もなく、両端キューをインポートする必要もありません)。
class Node:
def __init__(self, value = None):
self.left = None
self.right = None
self.value = value
ツリーの例を次に示します。
n1 = Node(1)
n2 = Node(2)
n3 = Node(3)
n1.left = n2
n1.right = n3
この例では、n1は、n2、n3を子として持つツリーのルートです。
もう少し「Pythonic」?
class Node:
def __init__(self, value):
self.value = value
self.left = None
self.right = None
def __repr__(self):
return str(self.value)
class BST:
def __init__(self):
self.root = None
def __repr__(self):
self.sorted = []
self.get_inorder(self.root)
return str(self.sorted)
def get_inorder(self, node):
if node:
self.get_inorder(node.left)
self.sorted.append(str(node.value))
self.get_inorder(node.right)
def add(self, value):
if not self.root:
self.root = Node(value)
else:
self._add(self.root, value)
def _add(self, node, value):
if value <= node.value:
if node.left:
self._add(node.left, value)
else:
node.left = Node(value)
else:
if node.right:
self._add(node.right, value)
else:
node.right = Node(value)
from random import randint
bst = BST()
for i in range(100):
bst.add(randint(1, 1000))
print (bst)
#!/usr/bin/python
class BinaryTree:
def __init__(self, left, right, data):
self.left = left
self.right = right
self.data = data
def pre_order_traversal(root):
print(root.data, end=' ')
if root.left != None:
pre_order_traversal(root.left)
if root.right != None:
pre_order_traversal(root.right)
def in_order_traversal(root):
if root.left != None:
in_order_traversal(root.left)
print(root.data, end=' ')
if root.right != None:
in_order_traversal(root.right)
def post_order_traversal(root):
if root.left != None:
post_order_traversal(root.left)
if root.right != None:
post_order_traversal(root.right)
print(root.data, end=' ')
私は多くの良い解決策がすでに投稿されていることを知っていますが、私は通常バイナリツリーに異なるアプローチを持っています:Nodeクラスを使用して直接実装する方が読みやすいですが、多くのノードがある場合は非常になりますメモリに貪欲なので、1つの複雑さの層を追加し、pythonリストにノードを保存し、リストのみを使用してツリーの動作をシミュレートすることをお勧めします。
Nodeクラスを定義して、必要に応じてツリー内のノードを最終的に表すこともできますが、リスト内の単純な形式[値、左、右]を維持すると、メモリの半分以下を使用できます!
以下は、ノードを配列に格納するバイナリ検索ツリークラスの簡単な例です。追加、削除、検索などの基本的な機能を提供します...
"""
Basic Binary Search Tree class without recursion...
"""
__author__ = "@fbparis"
class Node(object):
__slots__ = "value", "parent", "left", "right"
def __init__(self, value, parent=None, left=None, right=None):
self.value = value
self.parent = parent
self.left = left
self.right = right
def __repr__(self):
return "<%s object at %s: parent=%s, left=%s, right=%s, value=%s>" % (self.__class__.__name__, hex(id(self)), self.parent, self.left, self.right, self.value)
class BinarySearchTree(object):
__slots__ = "_tree"
def __init__(self, *args):
self._tree = []
if args:
for x in args[0]:
self.add(x)
def __len__(self):
return len(self._tree)
def __repr__(self):
return "<%s object at %s with %d nodes>" % (self.__class__.__name__, hex(id(self)), len(self))
def __str__(self, nodes=None, level=0):
ret = ""
if nodes is None:
if len(self):
nodes = [0]
else:
nodes = []
for node in nodes:
if node is None:
continue
ret += "-" * level + " %s\n" % self._tree[node][0]
ret += self.__str__(self._tree[node][2:4], level + 1)
if level == 0:
ret = ret.strip()
return ret
def __contains__(self, value):
if len(self):
node_index = 0
while self._tree[node_index][0] != value:
if value < self._tree[node_index][0]:
node_index = self._tree[node_index][2]
else:
node_index = self._tree[node_index][3]
if node_index is None:
return False
return True
return False
def __eq__(self, other):
return self._tree == other._tree
def add(self, value):
if len(self):
node_index = 0
while self._tree[node_index][0] != value:
if value < self._tree[node_index][0]:
b = self._tree[node_index][2]
k = 2
else:
b = self._tree[node_index][3]
k = 3
if b is None:
self._tree[node_index][k] = len(self)
self._tree.append([value, node_index, None, None])
break
node_index = b
else:
self._tree.append([value, None, None, None])
def remove(self, value):
if len(self):
node_index = 0
while self._tree[node_index][0] != value:
if value < self._tree[node_index][0]:
node_index = self._tree[node_index][2]
else:
node_index = self._tree[node_index][3]
if node_index is None:
raise KeyError
if self._tree[node_index][2] is not None:
b, d = 2, 3
Elif self._tree[node_index][3] is not None:
b, d = 3, 2
else:
i = node_index
b = None
if b is not None:
i = self._tree[node_index][b]
while self._tree[i][d] is not None:
i = self._tree[i][d]
p = self._tree[i][1]
b = self._tree[i][b]
if p == node_index:
self._tree[p][5-d] = b
else:
self._tree[p][d] = b
if b is not None:
self._tree[b][1] = p
self._tree[node_index][0] = self._tree[i][0]
else:
p = self._tree[i][1]
if p is not None:
if self._tree[p][2] == i:
self._tree[p][2] = None
else:
self._tree[p][3] = None
last = self._tree.pop()
n = len(self)
if i < n:
self._tree[i] = last[:]
if last[2] is not None:
self._tree[last[2]][1] = i
if last[3] is not None:
self._tree[last[3]][1] = i
if self._tree[last[1]][2] == n:
self._tree[last[1]][2] = i
else:
self._tree[last[1]][3] = i
else:
raise KeyError
def find(self, value):
if len(self):
node_index = 0
while self._tree[node_index][0] != value:
if value < self._tree[node_index][0]:
node_index = self._tree[node_index][2]
else:
node_index = self._tree[node_index][3]
if node_index is None:
return None
return Node(*self._tree[node_index])
return None
ノードを削除してBST構造を維持できるように、親属性を追加しました。
読みやすさ、特に「削除」機能については申し訳ありません。基本的に、ノードが削除されると、ツリー配列をポップして最後の要素に置き換えます(最後のノードを削除する場合を除く)。 BST構造を維持するために、削除されたノードは左の子の最大値または右の子の最小値に置き換えられ、インデックスを有効に保つためにいくつかの操作を行う必要がありますが、十分に高速です。
私はこのテクニックをより高度なものに使用して、内部基数トライを持ついくつかのビッグワード辞書を作成し、メモリ消費を7-8で割ることができました(例を参照できます: https://Gist.github .com/fbparis/b3ddd5673b603b42c880974b23db7cda )
import random
class TreeNode:
def __init__(self, key):
self.key = key
self.left = None
self.right = None
self.p = None
class BinaryTree:
def __init__(self):
self.root = None
def length(self):
return self.size
def inorder(self, node):
if node == None:
return None
else:
self.inorder(node.left)
print node.key,
self.inorder(node.right)
def search(self, k):
node = self.root
while node != None:
if node.key == k:
return node
if node.key > k:
node = node.left
else:
node = node.right
return None
def minimum(self, node):
x = None
while node.left != None:
x = node.left
node = node.left
return x
def maximum(self, node):
x = None
while node.right != None:
x = node.right
node = node.right
return x
def successor(self, node):
parent = None
if node.right != None:
return self.minimum(node.right)
parent = node.p
while parent != None and node == parent.right:
node = parent
parent = parent.p
return parent
def predecessor(self, node):
parent = None
if node.left != None:
return self.maximum(node.left)
parent = node.p
while parent != None and node == parent.left:
node = parent
parent = parent.p
return parent
def insert(self, k):
t = TreeNode(k)
parent = None
node = self.root
while node != None:
parent = node
if node.key > t.key:
node = node.left
else:
node = node.right
t.p = parent
if parent == None:
self.root = t
Elif t.key < parent.key:
parent.left = t
else:
parent.right = t
return t
def delete(self, node):
if node.left == None:
self.transplant(node, node.right)
Elif node.right == None:
self.transplant(node, node.left)
else:
succ = self.minimum(node.right)
if succ.p != node:
self.transplant(succ, succ.right)
succ.right = node.right
succ.right.p = succ
self.transplant(node, succ)
succ.left = node.left
succ.left.p = succ
def transplant(self, node, newnode):
if node.p == None:
self.root = newnode
Elif node == node.p.left:
node.p.left = newnode
else:
node.p.right = newnode
if newnode != None:
newnode.p = node.p
接続されたノードのNode
ベースのクラスは、標準的なアプローチです。これらは視覚化が難しい場合があります。
Python Patterns-Implementing Graphs の essay から動機付けられた、簡単な辞書を考えてみましょう。
Given
二分木
a
/ \
b c
/ \ \
d e f
コード
unique ノードの辞書を作成します。
tree = {
"a": ["b", "c"],
"b": ["d", "e"],
"c": [None, "f"],
"d": [None, None],
"e": [None, None],
"f": [None, None],
}
詳細
find_all_paths()
を参照)。ツリーベースの機能には、多くの場合、次の一般的な操作が含まれます。
これらすべての操作を実装してみてください。ここでは、これらの関数の one を示します-BFSトラバーサル:
例
import collections as ct
def traverse(tree):
"""Yield nodes from a tree via BFS."""
q = ct.deque() # 1
root = next(iter(tree)) # 2
q.append(root)
while q:
node = q.popleft()
children = filter(None, tree.get(node))
for n in children: # 3
q.append(n)
yield node
list(traverse(tree))
# ['a', 'b', 'c', 'd', 'e', 'f']
これは、ノードと子の辞書に適用される 幅優先検索(レベル順)アルゴリズム です。
deque
を使用しますが、 queue
またはlist
は機能します(後者は非効率的です)。木に関するこの詳細な tutorial も参照してください。
Insight
一般にトラバーサルについて素晴らしいことは、キューを stack に置き換えるだけで、 depth-first search(DFS) への後者の反復アプローチを簡単に変更できることです。別名LIFO Queue)。これは単に、エンキューした側と同じ側からデキューすることを意味します。 DFSを使用すると、各ブランチを検索できます。
どうやって? deque
を使用しているため、node = q.popleft()
をnode = q.pop()
に変更してスタックをエミュレートできます(右)。結果は、右向きの pre-ordered DFS :['a', 'c', 'f', 'b', 'e', 'd']
です。
この実装は、ツリーの構造を破壊することなく、挿入、検索、および削除の操作をサポートします。これはバランサードツリーではありません。
# Class for construct the nodes of the tree. (Subtrees)
class Node:
def __init__(self, key, parent_node = None):
self.left = None
self.right = None
self.key = key
if parent_node == None:
self.parent = self
else:
self.parent = parent_node
# Class with the structure of the tree.
# This Tree is not balanced.
class Tree:
def __init__(self):
self.root = None
# Insert a single element
def insert(self, x):
if(self.root == None):
self.root = Node(x)
else:
self._insert(x, self.root)
def _insert(self, x, node):
if(x < node.key):
if(node.left == None):
node.left = Node(x, node)
else:
self._insert(x, node.left)
else:
if(node.right == None):
node.right = Node(x, node)
else:
self._insert(x, node.right)
# Given a element, return a node in the tree with key x.
def find(self, x):
if(self.root == None):
return None
else:
return self._find(x, self.root)
def _find(self, x, node):
if(x == node.key):
return node
Elif(x < node.key):
if(node.left == None):
return None
else:
return self._find(x, node.left)
Elif(x > node.key):
if(node.right == None):
return None
else:
return self._find(x, node.right)
# Given a node, return the node in the tree with the next largest element.
def next(self, node):
if node.right != None:
return self._left_descendant(node.right)
else:
return self._right_ancestor(node)
def _left_descendant(self, node):
if node.left == None:
return node
else:
return self._left_descendant(node.left)
def _right_ancestor(self, node):
if node.key <= node.parent.key:
return node.parent
else:
return self._right_ancestor(node.parent)
# Delete an element of the tree
def delete(self, x):
node = self.find(x)
if node == None:
print(x, "isn't in the tree")
else:
if node.right == None:
if node.left == None:
if node.key < node.parent.key:
node.parent.left = None
del node # Clean garbage
else:
node.parent.right = None
del Node # Clean garbage
else:
node.key = node.left.key
node.left = None
else:
x = self.next(node)
node.key = x.key
x = None
# tests
t = Tree()
t.insert(5)
t.insert(8)
t.insert(3)
t.insert(4)
t.insert(6)
t.insert(2)
t.delete(8)
t.delete(5)
t.insert(9)
t.insert(1)
t.delete(2)
t.delete(100)
# Remember: Find method return the node object.
# To return a number use t.find(nº).key
# But it will cause an error if the number is not in the tree.
print(t.find(5))
print(t.find(8))
print(t.find(4))
print(t.find(6))
print(t.find(9))
here から取得したバイナリsearchツリーの適切な実装
'''
A binary search Tree
'''
from __future__ import print_function
class Node:
def __init__(self, label, parent):
self.label = label
self.left = None
self.right = None
#Added in order to delete a node easier
self.parent = parent
def getLabel(self):
return self.label
def setLabel(self, label):
self.label = label
def getLeft(self):
return self.left
def setLeft(self, left):
self.left = left
def getRight(self):
return self.right
def setRight(self, right):
self.right = right
def getParent(self):
return self.parent
def setParent(self, parent):
self.parent = parent
class BinarySearchTree:
def __init__(self):
self.root = None
def insert(self, label):
# Create a new Node
new_node = Node(label, None)
# If Tree is empty
if self.empty():
self.root = new_node
else:
#If Tree is not empty
curr_node = self.root
#While we don't get to a leaf
while curr_node is not None:
#We keep reference of the parent node
parent_node = curr_node
#If node label is less than current node
if new_node.getLabel() < curr_node.getLabel():
#We go left
curr_node = curr_node.getLeft()
else:
#Else we go right
curr_node = curr_node.getRight()
#We insert the new node in a leaf
if new_node.getLabel() < parent_node.getLabel():
parent_node.setLeft(new_node)
else:
parent_node.setRight(new_node)
#Set parent to the new node
new_node.setParent(parent_node)
def delete(self, label):
if (not self.empty()):
#Look for the node with that label
node = self.getNode(label)
#If the node exists
if(node is not None):
#If it has no children
if(node.getLeft() is None and node.getRight() is None):
self.__reassignNodes(node, None)
node = None
#Has only right children
Elif(node.getLeft() is None and node.getRight() is not None):
self.__reassignNodes(node, node.getRight())
#Has only left children
Elif(node.getLeft() is not None and node.getRight() is None):
self.__reassignNodes(node, node.getLeft())
#Has two children
else:
#Gets the max value of the left branch
tmpNode = self.getMax(node.getLeft())
#Deletes the tmpNode
self.delete(tmpNode.getLabel())
#Assigns the value to the node to delete and keesp tree structure
node.setLabel(tmpNode.getLabel())
def getNode(self, label):
curr_node = None
#If the tree is not empty
if(not self.empty()):
#Get tree root
curr_node = self.getRoot()
#While we don't find the node we look for
#I am using lazy evaluation here to avoid NoneType Attribute error
while curr_node is not None and curr_node.getLabel() is not label:
#If node label is less than current node
if label < curr_node.getLabel():
#We go left
curr_node = curr_node.getLeft()
else:
#Else we go right
curr_node = curr_node.getRight()
return curr_node
def getMax(self, root = None):
if(root is not None):
curr_node = root
else:
#We go deep on the right branch
curr_node = self.getRoot()
if(not self.empty()):
while(curr_node.getRight() is not None):
curr_node = curr_node.getRight()
return curr_node
def getMin(self, root = None):
if(root is not None):
curr_node = root
else:
#We go deep on the left branch
curr_node = self.getRoot()
if(not self.empty()):
curr_node = self.getRoot()
while(curr_node.getLeft() is not None):
curr_node = curr_node.getLeft()
return curr_node
def empty(self):
if self.root is None:
return True
return False
def __InOrderTraversal(self, curr_node):
nodeList = []
if curr_node is not None:
nodeList.insert(0, curr_node)
nodeList = nodeList + self.__InOrderTraversal(curr_node.getLeft())
nodeList = nodeList + self.__InOrderTraversal(curr_node.getRight())
return nodeList
def getRoot(self):
return self.root
def __isRightChildren(self, node):
if(node == node.getParent().getRight()):
return True
return False
def __reassignNodes(self, node, newChildren):
if(newChildren is not None):
newChildren.setParent(node.getParent())
if(node.getParent() is not None):
#If it is the Right Children
if(self.__isRightChildren(node)):
node.getParent().setRight(newChildren)
else:
#Else it is the left children
node.getParent().setLeft(newChildren)
#This function traversal the tree. By default it returns an
#In order traversal list. You can pass a function to traversal
#The tree as needed by client code
def traversalTree(self, traversalFunction = None, root = None):
if(traversalFunction is None):
#Returns a list of nodes in preOrder by default
return self.__InOrderTraversal(self.root)
else:
#Returns a list of nodes in the order that the users wants to
return traversalFunction(self.root)
#Returns an string of all the nodes labels in the list
#In Order Traversal
def __str__(self):
list = self.__InOrderTraversal(self.root)
str = ""
for x in list:
str = str + " " + x.getLabel().__str__()
return str
def InPreOrder(curr_node):
nodeList = []
if curr_node is not None:
nodeList = nodeList + InPreOrder(curr_node.getLeft())
nodeList.insert(0, curr_node.getLabel())
nodeList = nodeList + InPreOrder(curr_node.getRight())
return nodeList
def testBinarySearchTree():
r'''
Example
8
/ \
3 10
/ \ \
1 6 14
/ \ /
4 7 13
'''
r'''
Example After Deletion
7
/ \
1 4
'''
t = BinarySearchTree()
t.insert(8)
t.insert(3)
t.insert(6)
t.insert(1)
t.insert(10)
t.insert(14)
t.insert(13)
t.insert(4)
t.insert(7)
#Prints all the elements of the list in order traversal
print(t.__str__())
if(t.getNode(6) is not None):
print("The label 6 exists")
else:
print("The label 6 doesn't exist")
if(t.getNode(-1) is not None):
print("The label -1 exists")
else:
print("The label -1 doesn't exist")
if(not t.empty()):
print(("Max Value: ", t.getMax().getLabel()))
print(("Min Value: ", t.getMin().getLabel()))
t.delete(13)
t.delete(10)
t.delete(8)
t.delete(3)
t.delete(6)
t.delete(14)
#Gets all the elements of the tree In pre order
#And it prints them
list = t.traversalTree(InPreOrder, t.root)
for x in list:
print(x)
if __== "__main__":
testBinarySearchTree()