版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/m0_37324740/article/details/82763901
博主曾在之前的博客中用 python实现树结构,此篇博客将专门用 python 实现树的各种递归和非递归的遍历。
1. 构建树
我们先构建一棵简单的树:
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
a = TreeNode(1)
b = TreeNode(2)
c = TreeNode(3)
d = TreeNode(4)
e = TreeNode(5)
f = TreeNode(6)
g = TreeNode(7)
a.left = b
a.right = c
b.left = d
b.right = e
c.left = f
c.right = g
这是一棵这样的树:
2. 前序遍历
根节点->左子树->右子树
# 先序打印二叉树(递归)
def preOrderTraverse(node):
if not node:
return None
print(node.val)
preOrderTraverse(node.left)
preOrderTraverse(node.right)
# 先序打印二叉树(非递归)
def preOrderTravese(node):
stack = [node]
while len(stack) > 0:
print(node.val)
if node.right:
stack.append(node.right)
if node.left:
stack.append(node.left)
node = stack.pop()
输出:1245367
2. 中序遍历
左子树->根节点->右子树
# 中序打印二叉树(递归)
def inOrderTraverse(node):
if node is None:
return None
inOrderTraverse(node.left)
print(node.val)
inOrderTraverse(node.right)
# 中序打印二叉树(非递归)
def inOrderTraverse(node):
stack = []
pos = node
while pos or len(stack) > 0:
if pos:
stack.append(pos)
pos = pos.left
else:
pos = stack.pop()
print(pos.val)
pos = pos.right
输出:4251637
4. 后序遍历
左子树->右子树->根节点
# 后序打印二叉树(递归)
def postOrderTraverse(node):
if node is None:
return None
postOrderTraverse(node.left)
postOrderTraverse(node.right)
print(node.val)
# 后序打印二叉树(非递归)
# 使用两个栈结构
# 第一个栈进栈顺序:左节点->右节点->跟节点
# 第一个栈弹出顺序: 跟节点->右节点->左节点(先序遍历栈弹出顺序:跟->左->右)
# 第二个栈存储为第一个栈的每个弹出依次进栈
# 最后第二个栈依次出栈
def postOrderTraverse(node):
stack = [node]
stack2 = []
while len(stack) > 0:
node = stack.pop()
stack2.append(node)
if node.left is not None:
stack.append(node.left)
if node.right is not None:
stack.append(node.right)
while len(stack2) > 0:
print(stack2.pop().val)
输出:4526731
5. 层次遍历
逐层遍历
def layerTraverse(node):
if not node:
return None
queue = []
queue.append(node)
while len(queue) > 0:
tmp = queue.pop(0)
print(tmp.val)
if node.left:
queue.append(tmp.left)
if node.right:
queue.append(tmp.right)
输出:1234567