You are given the root
of a binary search tree (BST), where the values of exactly two nodes of the tree were swapped by mistake. Recover the tree without changing its structure.
Example 1:
Input: root = [1,3,null,null,2]
Output: [3,1,null,null,2]
Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid.
Example 2:
Input: root = [3,1,4,null,null,2]
Output: [2,1,4,null,null,3]
Explanation: 2 cannot be in the right subtree of 3 because 2 < 3. Swapping 2 and 3 makes the BST valid.
Constraints:
- The number of nodes in the tree is in the range
[2, 1000]
. -
-231 <= Node.val <= 231 - 1
Follow up: A solution using O(n)
space is pretty straight-forward. Could you devise a constant O(1)
space solution?SOLUTION:
# 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 inorder(self, root, append):
if root:
self.inorder(root.left, append)
if append:
self.vals.append(root.val)
else:
root.val = self.vals[self.i]
self.i += 1
self.inorder(root.right, append)
def recoverTree(self, root: Optional[TreeNode]) -> None:
self.vals = []
self.inorder(root, True)
self.vals.sort()
self.i = 0
self.inorder(root, False)
Top comments (0)