主要是：前序遍历、中序遍历、后序遍历、层级遍历、非递归前序遍历、非递归中序遍历、非递归后序遍历
复制代码 代码如下:
#!/usr/bin/env python
#-*- coding:utf8 -*-
class TreeNode(object):
def __init__(self, data=None, left=None, right=None):
self.data = data
self.left = left
self.right = right
class Tree(object):
def __init__(self, root=None):
self.root = None
def makeTree(self, data, left, right):
self.root = TreeNode(data, left, right)
def is_empty(self):
\”\”\”是否为空 \”\”\”
if self.root is None:
return True
return False
def preOrder(self, r):
\”\”\”前序遍历 \”\”\”
if not r.is_empty():
print r.root.data
if r.root.left is not None:
r.preOrder(r.root.left)
if r.root.right is not None:
r.preOrder(r.root.right)
def inOrder(self, r):
\”\”\”中序遍历 \”\”\”
if not r.is_empty():
if r.root.left is not None:
r.preOrder(r.root.left)
print r.root.data
if r.root.right is not None:
r.preOrder(r.root.right)
def postOrder(self, r):
\”\”\”后续遍历 \”\”\”
if not r.is_empty():
if r.root.left is not None:
r.preOrder(r.root.left)
print r.root.data
if r.root.right is not None:
r.preOrder(r.root.right)
def levelOrder(self, r):
\”\”\”层级遍历 \”\”\”
if not r.is_empty():
s = [r]
while len(s) > 0:
temp = s.pop(0) # 先弹出最先append到的点
if temp and temp.root is not None:
print temp.root.data
if temp.root.left is not None:
s.append(temp.root.left)
if self.root.right is not None:
s.append(temp.root.right)
def preOrder1(self, r):
\”\”\”非递归 前序遍历 \”\”\”
stack = []
current = r
while len(stack) > 0 or (current and not current.is_empty()):
while current and not current.is_empty():
print current.root.data
stack.append(current)
current = current.root.left
if len(stack) > 0:
current = stack.pop()
current = current.root.right
def inOrder1(self, r):
\”\”\”非递归 中序遍历 \”\”\”
stack = []
current = r
while len(stack) > 0 or (current and not current.is_empty()):
while current and not current.is_empty():
stack.append(current)
current = current.root.left
if len(stack) > 0:
current = stack.pop()
print current.root.data
current = current.root.right
def postOrder1(self, r):
\”\”\”非递归 后续遍历 \”\”\”
stack = []
current = r
pre = None
while len(stack) > 0 or (current and not current.is_empty()):
if current and not current.is_empty():
stack.append(current)
current = current.root.left
elif stack[-1].root.right != pre:
current = stack[-1].root.right
pre = None
else:
pre = stack.pop()
print pre.root.data
def leaves_count(self, r):
\”\”\”求叶子节点个数 \”\”\”
if r.is_empty():
return 0
elif (not r.root.left) and (not r.root.right):
return 1
else:
return r.root.left.leaves_count(r.root.left) + r.root.right.leaves_count(r.root.right)
if __name__ == \’__main__\’:
\”\”\”二叉树\”\”\”
ra, rb, rc, rd, re, rf = Tree(), Tree(), Tree(), Tree(), Tree(), Tree()
ra.makeTree(\”a\”, None, None)
rb.makeTree(\”b\”, None, None)
rc.makeTree(\”c\”, None, None)
rd.makeTree(\”d\”, None, None)
re.makeTree(\”e\”, None, None)
rf.makeTree(\”f\”, None, None)
r1, r2, r3, r4, r = Tree(), Tree(), Tree(), Tree(), Tree()
r1.makeTree(\”-\”, rc, rd)
r2.makeTree(\”*\”, rb, r1)
r3.makeTree(\”+\”, ra, r2)
r4.makeTree(\”/\”, re, rf)
r.makeTree(\”-\”, r3, r4)
r.preOrder(r)
r.inOrder(r)
r.postOrder(r)
r.levelOrder(r)
print r.leaves_count(r)
大学的时候学过kmp算法，最近在看的时候发现竟然忘了，所以去重新看了看书，然后用python写下了这个算法：
复制代码 代码如下:
def kmp(text, pattern):
\”\”\”kmp算法 \”\”\”
pattern = list(pattern)
next = [-1] * len(pattern)
#next 函数
i, j = 1, -1
for i in range(1, len(pattern)):
j = next[i – 1]
while True:
if pattern[i – 1] == pattern[j] or j == -1:
next[i] = j + 1
break
else:
j = next[j]
#循环比较
i, j = 0, 0
while i < len(text) and j < len(pattern):
if text[i] == pattern[j] or j == -1:
i += 1
j += 1
else:
j = next[j]
#返回结果 如果匹配，返回匹配的位置，否则返回-1
if j == len(pattern):
print i – j
else:
print -1