首页 > 代码库 > [LeetCode]94. Binary Tree Inorder Traversal

[LeetCode]94. Binary Tree Inorder Traversal

Given a binary tree, return the inorder traversal of its nodes‘ values.

For example:
Given binary tree [1,null,2,3],

   1
         2
    /
   3

 

return [1,3,2].

题意:中序遍历树

先写一种比较蠢的方法,我可以借口对python的数据结构不熟悉

 1 class Solution(object):
 2     def inorderTraversal(self, root):
 3         """
 4         :type root: TreeNode
 5         :rtype: List[int]
 6         """
 7         res = []
 8         if root != None:
 9             if root.left!=None:
10                 for i in self.inorderTraversal(root.left):
11                     res.append(i)
12             res.append(root.val)
13             if root.right!=None:
14                 for i in self.inorderTraversal(root.right):
15                     res.append(i)
16         return res

一种正常的写法:

 1 class Solution(object):
 2     def inorderTraversal(self, root):
 3         """
 4         :type root: TreeNode
 5         :rtype: List[int]
 6         """
 7         res = []
 8         self.helper(root,res)
 9         return res
10         
11         
12     def helper(self,root,res):
13         if root:
14             self.helper(root.left,res)
15             res.append(root.val)
16             self.helper(root.right,

上面两种方法都使用了递归,下面一种未使用递归的方法:

 1 class Solution(object):
 2     def inorderTraversal(self, root):
 3         """
 4         :type root: TreeNode
 5         :rtype: List[int]
 6         """
 7         res,s = [],[]
 8         while True:
 9             while root:
10                 s.append(root)
11                 root = root.left
12             if not s:
13                 return res
14             node = s.pop()
15             res.append(node.val)
16             root=node.right

顺便吐槽一句:为什么把这个题分到哈希类???!!!

[LeetCode]94. Binary Tree Inorder Traversal