Morris主要解決一個問題,如何實現二叉樹的前中後序遍歷,有兩個要求:
1. O(1)空間複雜度,O(N)時間複雜度;
2. 二叉樹的形狀不能被破壞(中間過程允許改變其形狀)。
以往是用棧結構輔助實現從下到上,但此算法是用到了線索二叉樹的概念,通過讓底層結點指向null的空閒指針指向上層的某個結點。
先不論先中後序,Morris遍歷的過程爲:
1、如果cur 爲null,則過程停止,否則繼續;
2、如果cur沒有左子樹,則cur = cur.right;
3、如果cur有左子樹,則找到cur左子樹上最右的結點,記爲mostRight
1)如果mostRight的right指針指向null,則mostRight.right = cur,然後cur = cur.left
2)如果mostRight的right指針指向cur(即已經經過第一次遍歷),令mostRight.right = null,cur = cur.right
根據以上過程寫出的代碼爲:
def Morris(self,root):
if not root:
return None
cur = root
while cur:
mostRight = cur.left
if mostRight:
while mostRight.right and mostRight.right != cur:
mostRight = mostRight.right
if mostRight.right == None:
mostRight.right = cur
cur = cur.left
continue
else:
mostRight.right = None
cur = cur.right
那麼先序遍歷,則是將打印函數放在第一次遍歷到的結點處:
def Morris(self,root):
if not root:
return None
cur = root
while cur:
mostRight = cur.left
if mostRight:
while mostRight.right and mostRight.right != cur:
mostRight = mostRight.right
if mostRight.right == None:
mostRight.right = cur
print(cur.val)#這裏是根結點
cur = cur.left
continue
else:
mostRight.right = None
else:
print(cur.val)#無左就直接打印右
cur = cur.right
中序遍歷,則是等左子樹都遍歷完成後再開始打印
def Morris(self,root):
if not root:
return None
cur = root
while cur:
mostRight = cur.left
if mostRight:
while mostRight.right and mostRight.right != cur:
mostRight = mostRight.right
if mostRight.right == None:
mostRight.right = cur
cur = cur.left
continue
else:
mostRight.right = None
print(cur.val)
cur = cur.right
後序比較複雜,程序如下:
1、對於無左子樹的(只遍歷一次的),直接跳過,無打印行爲;
2、對於有左子樹的(可以遍歷兩次的),cur第一次時無打印,第二次時逆序打印左子樹的右邊界;
3、cur遍歷完成後,逆序打印整個樹的右邊界。
其實整個程序就如下圖所示:
打印順序爲4,5,2,6,7,3,1
def Morris(self,root):
if not root:
return None
cur = root
while cur:
mostRight = cur.left
if mostRight:#有左子樹,可以遍歷兩次
while mostRight.right and mostRight.right != cur:
mostRight = mostRight.right
if mostRight.right == None:
mostRight.right = cur
cur = cur.left
continue
else:
mostRight.right = None
#第二次
self.printEdge(cur.left)
cur = cur.right
# 最後逆序打印整棵樹右邊界
self.printEdge(root)
def printEdge(self,head):
tail = self.reverseEdge(head)
cur = tail
while cur:
print(cur.val)
cur = cur.right
#不能改變二叉樹結構,逆序完了逆回去
self.reverseEdge(tail)
def reverseEdge(self,cur):
pre = None
while cur:
next = cur.right
cur.right = pre
pre = cur
cur = next
return pre
最後證明一下Morris符合題目要求:
代碼只使用了有限幾個變量,故額外空間複雜度爲O(1),而遍歷對於有左子樹的只遍歷兩次,無左子樹的只遍歷一次,總時間複雜度還是O(N)