LeetCode 94 Binary Tree Inorder Traversal(Python詳解及實現)
【題目】
Given a binary tree, return the inordertraversal of its nodes' values.
For example:
Given binary tree [1,null,2,3],
1
\
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, couldyou do it iteratively?
給定一個二叉樹返回其中序遍歷
注意:遞迴解法過於簡單,你能用迭代的方式嗎?
【思路】
二叉樹遍歷:
前序遍歷:根結點、左子樹、右子樹
中序遍歷:左子樹、根結點、右子樹
後序遍歷:左子樹、右子樹、根結點
如下一棵二叉樹,利用棧其整個過程為:
l 根結點G入棧,若入棧的結點存在左子樹,則依次入棧G/D/A,直至A發現其左子樹為空,停止入棧,此時棧stack = [G,D,A]。
l A出棧,並對A進行遍歷,發現A沒有右子樹,根據中序遍歷,需要遍歷A的根節點D,D出棧,D存在右孩子,將其右孩子F入棧,F有左子樹E,E入棧,此時stack = [G,F,E],res=[A,D]
l E出棧,並對E進行遍歷,發現沒有右子樹,根據中序遍歷,需要遍歷E的根結點F,F出棧,此時棧為stack = [G],res = [A,D, E, F]
l G出棧,並遍歷G,G有右子樹,將右子樹M入棧,此時棧為stack = [M],res = [A,D, E, F,G]
l 右子樹根結點M,按照中序遍歷規則重複以上步驟:
M存在左子樹H入棧,stack = [M,H],res= [A,D, E, F,G]
H出棧,遍歷H,發現H沒有右子樹,根據中序遍歷,需要遍歷H的根結點M,M有右子樹Z,Z入棧,此時stack = [Z],res = [A,D, E, F,G,H,M]
Z出棧,遍歷Z,發現Z沒有右子樹,此時stack= [],res = [A,D, E, F,G,H,M,Z]
【Python實現】
方法一:非遞迴方式
#輸出二叉樹的中序遍歷,非遞迴方式
# Definition for a binary tree node. class TreeNode(object): def __init__(self, x): self.val = x self.left = None self.right = None class Solution(object): def inorderTraversal(self, root): """ :type root: TreeNode :rtype: List[int] """ res = [] self.iterative_inorder(root, res) print(res) return res def iterative_inorder(self, root, res):#迭代中序遍歷 stack = [] while root or stack: if root: stack.append(root) root = root.left else: root = stack.pop() res.append(root.val) root = root.right return res if __name__ == '__main__': S = Solution() l1 = TreeNode(1) l2 = TreeNode(2) l3 = TreeNode(3) l4 = TreeNode(4) l5 = TreeNode(5) l6 = TreeNode(6) l7 = TreeNode(7) root = l1 l1.left = l2 l1.right = l3 l2.left = l4 l2.right = l5 l3.left = l6 l3.right = l7 S.inorderTraversal(root)
方法二:遞迴方式
# Definition for a binary tree node.
class TreeNode(object):
def __init__(self, x):
self.val = x
self.left = None
self.right = None
class Solution(object):
def inorderTraversal(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
res = []
self. recursive_inorder (root, res)
print(res)
return res
def recursive_inorder(self, root, res):#遞迴中序遍歷
if root:
self.recursive_inorder(root.left, res)
res.append(root.val)
self.recursive_inorder(root.right, res)
if __name__ == '__main__':
S = Solution()
l1 = TreeNode(1)
l2 = TreeNode(2)
l3 = TreeNode(3)
l4 = TreeNode(4)
l5 = TreeNode(5)
l6 = TreeNode(6)
l7 = TreeNode(7)
root = l1
l1.left = l2
l1.right = l3
l2.left = l4
l2.right = l5
l3.left = l6
l3.right = l7
S.inorderTraversal(root)