【python/hard/99】Recover Binary Search Tree

题目

https://leetcode.com/problems/recover-binary-search-tree/description/
打开链接吧,今天懒得粘了,题目有点长

基本思路

BST ==》中序遍历是有序的
如果被误操作:
BST ==》中序遍历一定不是有序的
找到就好啦~
第一个乱序的数字是pre,第二个乱序的数字是root,用两个指针分别保存,然后交换值即可,不需要把树完全遍历完。

时间复杂度 O(n)
空间复杂度 O(1)

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def recoverTree(self, root):
        """
        :type root: TreeNode
        :rtype: void Do not return anything, modify root in-place instead.
        """
        self.n1 = self.n2 = None
        self.prev = None
        self.findTwoNodes(root)
        self.n1.val,self.n2.val = self.n2.val,self.n1.val

    
    def findTwoNodes(self,root):
        if root:
            self.findTwoNodes(root.left)
            if self.prev and self.prev.val > root.val:
                if self.n1 == None:
                    self.n1 = self.prev
                self.n2 = root
            self.prev = root
            self.findTwoNodes(root.right)
        

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