897. Increasing Order Search Tree

题目来源:  
 https://leetcode.com/problems/increasing-order-search-tree/
自我感觉难度/真实难度:medium/easy
题意:
 
分析:
 
自己的代码:
 
优秀代码:
 
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def increasingBST(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        array=self.inOrder(root)
        if not array:
            return 
        newRoot=TreeNode(array[0])
        curr=newRoot
        for i in range(1,len(array)):
            curr.right=TreeNode(array[i])
            curr=curr.right
        return newRoot
    
    
    def inOrder(self,root):
        
        
        if not root:
            return []
        res=[]
        res.extend(self.inOrder(root.left))
        res.append(root.val)
        res.extend(self.inOrder(root.right))
        return res

代码效率/结果:
        Runtime: 200 ms, faster than 45.33% of Python3 online submissions forIncreasing Order Search Tree.
class Solution:
  def increasingBST(self, root):
    dummy = TreeNode(0)
    self.prev = dummy
    def inorder(root):
      if not root: return None
      inorder(root.left)
      root.left = None
      self.prev.right = root
      self.prev = root
      inorder(root.right)
    inorder(root)
    return dummy.right

      Runtime: 160 ms, faster than 63.16% of Python3 online submissions forIncreasing Order Search Tree.

自己优化后的代码:
 

反思改进策略:

            1.前序遍历不熟悉,需要熟练编写这个代码

    2.看不懂优化的第二个解答: .prev     TreeNode(0)

原文地址:https://www.cnblogs.com/captain-dl/p/10165865.html