99-recover-binary-search-tree

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Description

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(Python)

# 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.
        """
        self.better(root)

    # Time Complexity: O(n)
    # Space Complexity: O(n)
    def naive(self, root):
        def traversal(node):
            return (
                traversal(node.left) + [node.val] + traversal(node.right)
                if node
                else []
            )

        inorder = traversal(root)

        n = len(inorder)

        for i in range(n):
            key = inorder[i]
            j = i - 1
            while j >= 0 and inorder[j] >= key:
                inorder[j + 1] = inorder[j]
                j -= 1
            inorder[j + 1] = key

        def update(node):
            nonlocal i
            if node:
                update(node.left)
                node.val = inorder[i]
                i += 1
                update(node.right)

        i = 0
        update(root)

    # Time Complexity: O(n)
    # Space Complexity: O(1)
    def better(self, root):
        cur, prev, First, Second = root, TreeNode(-float("inf")), None, None
        i = 0

        while cur:

            if cur.left:
                pre = cur.left
                while pre.right and pre.right != cur:
                    pre = pre.right

                if pre.right is None:
                    pre.right = cur
                    cur = cur.left
                else:
                    if cur.val < prev.val:
                        if First is None:
                            First = prev
                        Second = cur
                    pre.right = None
                    prev = cur
                    cur = cur.right
            else:
                if cur.val < prev.val:
                    if First is None:
                        First = prev
                    Second = cur
                prev = cur
                cur = cur.right
        First.val, Second.val = Second.val, First.val