99. 恢复二叉搜索树

问题描述

https://leetcode.cn/problems/recover-binary-search-tree/description/

解题思路

二叉搜索树的中序遍历一定是一个有序的序列。

题目中说两个节点的值被交换了，即我们中序遍历得到的二叉搜索树是无序的，但交换两个节点即为有序。

所以，我们可以：

1. 深度优先搜索，搞到二叉搜索树的序列。
2. 将这个序列排序，再次遍历，找到需要交换值的节点。
3. 将值进行交换。

代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def recoverTree(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        li = []
        def dfs(root):
            if not root:
                return
            dfs(root.left)
            li.append((root.val, root))
            dfs(root.right)
        dfs(root)
        li_sort = sorted(li, key=lambda x: x[0])
        swaped = []
        for i in range(len(li)):
            if li[i][0] != li_sort[i][0]:
                swaped.append(li[i][1])
        x, y = swaped
        x.val, y.val = y.val, x.val

 2024二刷：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def recoverTree(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        mem = []
        vals = []
        def dfs(root):
            if root:
                dfs(root.left)
                mem.append(root)
                vals.append(root.val)
                dfs(root.right)
        dfs(root)
        vals.sort()
        for i in range(len(mem)):
            mem[i].val = vals[i]