• 分類：BackTracking

• 考察知識點(diǎn)：BackTracking 二叉樹(Binary Tree)

• 最優(yōu)解時(shí)間復(fù)雜度：**BackTracking:O(???) ** BackTracking不好求 **Stack:O(n) **

94. Binary Tree Inorder Traversal

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?

代碼：

解法1(BackTracking):

# 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=[]
        
        if root==None:
            return res
        self.Traversal(root,res)
        return res
        
    def Traversal(self,root,res):
        if root == None:
            return
        self.Traversal(root.left,res)
        res.append(root.val)
        self.Traversal(root.right,res)

解法2(Stack):

# 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=[]
        
        if root==None:
            return res
        
        stack=[]
        pointer=root
        while(pointer!=None or len(stack)!=0):
            while pointer!=None:
                stack.append(pointer)
                pointer=pointer.left
            pointer=stack.pop()
            res.append(pointer.val)
            pointer=pointer.right
        return res

討論：

1.考一個(gè)二叉樹的遍歷，左中右的遍歷
2.這道題有兩種方法做這個(gè)題，一個(gè)是用Stack,一個(gè)是用回溯

3. inorder traversal 記一下這個(gè)名詞是左中右遍歷

BackTracking

Stack