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Leetcode 94. Binary Tree Inorder Traversal

时间:2019-07-15 19:13:06      阅读:68      评论:0      收藏:0      [点我收藏+]

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

Example:

Input: [1,null,2,3]
   1
         2
    /
   3

Output: [1,3,2]

Follow up: Recursive solution is trivial, could you do it iteratively?


  • 中序遍历。复习下递归跟迭代,需要借助stack。
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 1 # Definition for a binary tree node.
 2 # class TreeNode:
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.left = None
 6 #         self.right = None
 7 
 8 class Solution:
 9     def helper(self, node: TreeNode, list_result: List[int]):
10         if node is None:
11             return
12         
13         self.helper(node.left, list_result)
14 
15         list_result.append(node.val)
16         
17         self.helper(node.right, list_result)
18         
19     # Recursive approach    
20     def inorderTraversal1(self, root: TreeNode) -> List[int]:
21         if root is None:
22             return [] # should return [] rather than None
23         
24         res = []
25         self.helper(root, res)
26         
27         return res        
28     
29     # Iterative approach
30     def inorderTraversal(self, root: TreeNode) -> List[int]:
31         if root is None:
32             return []
33         
34         cur = root
35         res, stack = [], []
36         
37         while cur or stack:
38             while cur:
39                 stack.append(cur)
40                 cur = cur.left
41                 
42             cur = stack.pop()
43             res.append(cur.val)
44             
45             cur = cur.right
46             
47         return res
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Leetcode 94. Binary Tree Inorder Traversal

原文:https://www.cnblogs.com/pegasus923/p/11190606.html

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