897. Increasing Order Search Tree

题目来源：  

 https://leetcode.com/problems/increasing-order-search-tree/

自我感受难度/真实难度：medium/easy

 

题意：

 

分析：

 

本身的代码：

 

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.

 

本身优化后的代码：

 

反思改进策略：

            1.前序遍历不熟悉，须要熟练编写这个代码

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