原題

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]

Reference Answer

思路分析

直接按照正常中序遍歷走就好了，遞迴做，省事。

My Code

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
 

#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        
        res = []
        self.helper(root, res)
        return res
        
        
    def
 
 helper(self, root, res):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root:
            return
        else:
            self.helper(root.left, res)
            res.append(root.val)
            self.helper(root.right, res)

        

Reference Code

Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        this.val = val
        this.left, this.right = None, None
"""

class Solution:
    """
    @param root: The root of binary tree.
    @return: Inorder in ArrayList which contains node values.
    """
    def inorderTraversal(self, root):
        if root is None:
            return []
        else:
            return self.inorderTraversal(root.left) + [root.val] \
                              + self.inorderTraversal(root.right)
  

Note:

• 直接用遞迴做吧，省事，參考答案的最省事，不過有一點要注意就是 [] 使用，以及 + 的使用，這個是直接返回一個list，如果是範圍[[], [], []...]，就不能再直接這樣使用。

參考資料

[1] https://www.kancloud.cn/kancloud/data-structure-and-algorithm-notes/73026