二叉树Python的各类实现
本文介绍了二叉树的各类实现，并附上python代码
二叉树的概念
二叉树（Binary Tree）是包含n个节点的有限集合，该集合或者为空集（此时，二叉树称为空树），或者由一个根节点和两棵互不相交的、分别称为根节点的左子树和右子树的二叉树组成。
二叉树遍历和迭代方法
该文中的概念和实现方法要求均来自LeetCode：
https://leetcode.com/explore/learn/card/data-structure-tree/
- 二叉树的前序，后序，中序的迭代法实现和堆栈法实现；
- 二叉树的层次遍历
- 计算二叉树的深度宽度；
- 查看二叉树是否含有某路径；
- 查看二叉树是否对称；
python代码
#coding=utf-8
import queue
class Node(object):
"""节点类"""
def __init__(self, elem=-1, lchild=None, rchild=None):
self.elem = elem
self.lchild = lchild
self.rchild = rchild
class Tree(object):
"""树类"""
def __init__(self):
self.root = Node()
self.myQueue = []
def add(self, elem):
"""为树添加节点"""
node = Node(elem)
if self.root.elem == -1: # 如果树是空的，则对根节点赋值
self.root = node
self.myQueue.append(self.root)
else:
treeNode = self.myQueue[0] # 此结点的子树还没有齐。
if treeNode.lchild == None:
treeNode.lchild = node
self.myQueue.append(treeNode.lchild)
else:
treeNode.rchild = node
self.myQueue.append(treeNode.rchild)
self.myQueue.pop(0) # 如果该结点存在右子树，将此结点丢弃。
def front_digui(self, root):
"""利用递归实现树的先序遍历"""
if root == None:
return
print(root.elem,)
self.front_digui(root.lchild)
self.front_digui(root.rchild)
def middle_digui(self, root):
"""利用递归实现树的中序遍历"""
if root == None:
return
self.middle_digui(root.lchild)
print(root.elem,)
self.middle_digui(root.rchild)
def later_digui(self, root):
"""利用递归实现树的后序遍历"""
if root == None:
return
self.later_digui(root.lchild)
self.later_digui(root.rchild)
print(root.elem,)
def front_stack(self, root):
"""利用堆栈实现树的先序遍历"""
if root == None:
return
myStack = []
node = root
while node or myStack:
while node: #从根节点开始，一直找它的左子树
print(node.elem,)
myStack.append(node)
node = node.lchild
node = myStack.pop() #while结束表示当前节点node为空，即前一个节点没有左子树了
node = node.rchild #开始查看它的右子树
def middle_stack(self, root):
"""利用堆栈实现树的中序遍历"""
if root == None:
return
myStack = []
node = root
while node or myStack:
while node: #从根节点开始，一直找它的左子树
myStack.append(node)
node = node.lchild
node = myStack.pop() #while结束表示当前节点node为空，即前一个节点没有左子树了
print(node.elem,)
node = node.rchild #开始查看它的右子树
def later_stack(self, root):
"""利用堆栈实现树的后序遍历"""
if root == None:
return
myStack1 = []
myStack2 = []
node = root
myStack1.append(node)
while myStack1: #这个while循环的功能是找出后序遍历的逆序，存在myStack2里面
node = myStack1.pop()
if node.lchild:
myStack1.append(node.lchild)
if node.rchild:
myStack1.append(node.rchild)
myStack2.append(node)
while myStack2: #将myStack2中的元素出栈，即为后序遍历次序
print(myStack2.pop().elem,)
def level_queue(self, root):
"""利用队列实现树的层次遍历"""
if root == None:
return
myQueue = []
node = root
myQueue.append(node)
while myQueue:
node = myQueue.pop(0)
print(node.elem,)
if node.lchild != None:
myQueue.append(node.lchild)
if node.rchild != None:
myQueue.append(node.rchild)
def levelOrder(self,root):
"""BFS层次"""
res = []
if root == None:
return res
q = [root]
while len(q) != 0:
res.append([node.elem for node in q])
new_q = []
for node in q:
if node.lchild:
new_q.append(node.lchild)
if node.rchild:
new_q.append(node.rchild)
q = new_q
return res
def treeDepth(self,tree):
if tree == None:
return 0
leftDepth = self.treeDepth(tree.lchild)
rightDepth = self.treeDepth(tree.rchild)
if leftDepth > rightDepth:
return leftDepth + 1
if rightDepth >= leftDepth:
return rightDepth + 1
def button_up_maxDepth(self, root):
if root == None:
return 0
leftmax = self.button_up_maxDepth(root.lchild)
rightmax = self.button_up_maxDepth(root.rchild)
return max(leftmax, rightmax) + 1
answer = 0
def top_down_maxDepth(self, root,deepth):
if root == None:
return
if root.lchild == None and root.rchild == None:
self.answer= max(self.answer,deepth)
#print(self.answer)
self.top_down_maxDepth(root.lchild,deepth+1)
self.top_down_maxDepth(root.rchild,deepth+1)
def treeWidth(self,tree):
curwidth = 1
maxwidth = 0
q = queue.Queue()
q.put(tree)
while not q.empty():
n = curwidth
for i in range(n):
tmp = q.get()
curwidth -= 1
if tmp.rchild:
q.put(tmp.lchild)
curwidth += 1
if tmp.rchild:
q.put(tmp.rchild)
curwidth += 1
if curwidth > maxwidth:
maxwidth = curwidth
return maxwidth
def help(self, p, q):
if p == None and q == None:
return True
if p and q and p.elem == q.elem:
return self.help(p.rchild, q.lchild) and self.help(p.lchild, q.rchild)#递归判断左子树根的右子树根节点和右子树根的左子树根节点以及左子树根的左子树根节点和右子树根的右子树根节点的值是否相等
return False
def isSymmetric(self, root):
if root:
return self.help(root.lchild, root.rchild)
return True
def hasPathSum(self, root, sum: int):
if root == None:
return False
if root.lchild == None and root.rchild == None:
return root.elem == sum
return self.hasPathSum(root.lchild,sum - root.elem) or self.hasPathSum(root.rchild, sum - root.elem)
if __name__ == '__main__':
"""主函数"""
elems = range(10) #生成十个数据作为树节点
tree = Tree() #新建一个树对象
for elem in elems:
tree.add(elem) #逐个添加树的节点
print('队列实现层次遍历:')
tree.level_queue(tree.root)
print('\n\n递归实现先序遍历:')
tree.front_digui(tree.root)
print('\n递归实现中序遍历:')
tree.middle_digui(tree.root)
print('\n递归实现后序遍历:')
tree.later_digui(tree.root)
print('\n\n堆栈实现先序遍历:')
tree.front_stack(tree.root)
print('\n堆栈实现中序遍历:')
tree.middle_stack(tree.root)
print('\n堆栈实现后序遍历:')
tree.later_stack(tree.root)
print('\n堆栈深度:')
print(tree.treeDepth(tree.root))
print('\n堆栈宽度:')
print(tree.treeWidth(tree.root))
print('\n堆栈分层排序:')
print(tree.levelOrder(tree.root))
print('\nbutton up堆栈深度:')
print(tree.button_up_maxDepth(tree.root))
print('\ntop down堆栈深度:')
tree.top_down_maxDepth(tree.root,1)
print(tree.answer)
print('\n是否有path sum等于11:')
print(tree.hasPathSum(tree.root,8))
elems2 = [1, 2, 2, 3, 4, 4, 3]
tree2 = Tree() # 新建一个树对象
for elem in elems2:
tree2.add(elem) # 逐个添加树的节点
print('\n堆栈分层排序:')
print(tree2.levelOrder(tree2.root))
print('\n是否是堆成二叉树:')
print(tree2.isSymmetric(tree2.root))
运行结果
binary_tree.py
队列实现层次遍历:
0
1
2
3
4
5
6
7
8
9
递归实现先序遍历:
0
1
3
7
8
4
9
2
5
6
递归实现中序遍历:
7
3
8
1
9
4
0
5
2
6
递归实现后序遍历:
7
8
3
9
4
1
5
6
2
0
堆栈实现先序遍历:
0
1
3
7
8
4
9
2
5
6
堆栈实现中序遍历:
7
3
8
1
9
4
0
5
2
6
堆栈实现后序遍历:
7
8
3
9
4
1
5
6
2
0
堆栈深度:
4
堆栈宽度:
4
堆栈分层排序:
[[0], [1, 2], [3, 4, 5, 6], [7, 8, 9]]
button up堆栈深度:
4
top down堆栈深度:
4
是否有path sum等于11:
True
堆栈分层排序:
[[1], [2, 2], [3, 4, 4, 3]]
是否是堆成二叉树:
True
Process finished with exit code 0