Leetcode #99: Recover Binary Search Tree

In this guide, we solve Leetcode #99 Recover Binary Search Tree in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

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.

Quick Facts

Difficulty: Medium
Premium: No
Tags: Tree, Depth-First Search, Binary Search Tree, Binary Tree

Intuition

We need to explore a structure deeply before backing up, which suits DFS.

DFS keeps local context on the call stack and is easy to implement recursively.

Approach

Define a recursive DFS that carries the necessary state.

Combine child results as the recursion unwinds.

Steps:

Define a recursive DFS with state.
Visit children and combine results.
Return the final aggregation.

Example

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.

Python 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 recoverTree(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """

        def dfs(root):
            if root is None:
                return
            nonlocal prev, first, second
            dfs(root.left)
            if prev and prev.val > root.val:
                if first is None:
                    first = prev
                second = root
            prev = root
            dfs(root.right)

        prev = first = second = None
        dfs(root)
        first.val, second.val = second.val, first.val

Complexity

The time complexity is $O(n)$ , and the space complexity is $O(n)$ . The space complexity is $O(n)$ .

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.

Python 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 recoverTree(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """

        def dfs(root):
            if root is None:
                return
            nonlocal prev, first, second
            dfs(root.left)
            if prev and prev.val > root.val:
                if first is None:
                    first = prev
                second = root
            prev = root
            dfs(root.right)

        prev = first = second = None
        dfs(root)
        first.val, second.val = second.val, first.val

Leetcode #99: Recover Binary Search Tree

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #99: Recover Binary Search Tree

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview