How to judge different types of binary trees using Python-Python Tutorial-php.cn

How to judge different types of binary trees using Python

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Release： 2024-01-23 10:06:06

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二叉树是一种树状数据结构，其中每个父节点最多可以有两个子节点。

二叉树的类型

完全二叉树

完全二叉树是一种特殊类型的二叉树，其父节点存在2种情况，要么有2个子节点，要么没有子节点，详情如下图：

完全二叉树定理

1、叶数为i+1

2、节点总数为2i+1

3、内部节点数为(n–1)/2

4、叶数为(n+1)/2

5、节点总数为2l–1

6、内部节点数为l–1

7、叶子的数量最多2^λ-1

Python判断完整二叉树

class Node:
def __init__(self,item):
self.item=item
self.leftChild=None
self.rightChild=None
def isFullTree(root):
if root is None:
return True
if root.leftChild is None and root.rightChild is None:
return True
if root.leftChild is not None and root.rightChild is not None:
return(isFullTree(root.leftChild)and isFullTree(root.rightChild))
return False
root=Node(1)
root.rightChild=Node(3)
root.leftChild=Node(2)
root.leftChild.leftChild=Node(4)
root.leftChild.rightChild=Node(5)
root.leftChild.rightChild.leftChild=Node(6)
root.leftChild.rightChild.rightChild=Node(7)
if isFullTree(root):
print("The tree is a full binary tree")
else:
print("The tree is not a full binary tree")

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完美二叉树

完美二叉树的每个内部节点都恰好有两个子节点，并且所有叶节点都在同一级别，如下图：

完美二叉树定理

1、高度为h的完美二叉树有2^(h+1)–1个节点

2、具有n个节点的完美二叉树的高度为log(n+1)–1=Θ(ln(n))。

3、高度为h的完美二叉树具有2^h节点

4、完美二叉树中节点的平均深度为Θ(ln(n))。

Python判断完美二叉树

class newNode:
def __init__(self,k):
self.key=k
self.right=self.left=None
def calculateDepth(node):
d=0
while(node is not None):
d+=1
node=node.left
return d
def is_perfect(root,d,level=0):
if(root is None):
return True
if(root.left is None and root.right is None):
return(d==level+1)
if(root.left is None or root.right is None):
return False
return(is_perfect(root.left,d,level+1)and
is_perfect(root.right,d,level+1))
root=None
root=newNode(1)
root.left=newNode(2)
root.right=newNode(3)
root.left.left=newNode(4)
root.left.right=newNode(5)
if(is_perfect(root,calculateDepth(root))):
print("The tree is a perfect binary tree")
else:
print("The tree is not a perfect binary tree")

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退化或病态树

退化或病态树只具有左或右单个子节点的二叉树，如下图：

斜二叉树

倾斜二叉树要么由左节点支配，要么由右节点支配。因此，有左二叉树和右二叉树两种类型，如下图：

平衡二叉树

平衡二叉树每个节点的左子树和右子树的高度之差为0或1，如下图：

Python判断平衡二叉树

class Node:
def __init__(self,data):
self.data=data
self.left=self.right=None
class Height:
def __init__(self):
self.height=0
def isHeightBalanced(root,height):
left_height=Height()
right_height=Height()
if root is None:
return True
l=isHeightBalanced(root.left,left_height)
r=isHeightBalanced(root.right,right_height)
height.height=max(left_height.height,right_height.height)+1
if abs(left_height.height-right_height.height)&lt;=1:
return l and r
return False
height=Height()
root=Node(1)
root.left=Node(2)
root.right=Node(3)
root.left.left=Node(4)
root.left.right=Node(5)
if isHeightBalanced(root,height):
print(&#x27;The tree is balanced&#x27;)
else:
print(&#x27;The tree is not balanced&#x27;)

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