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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0099_recover_binary_search_tree;

// #Medium #Depth_First_Search #Tree #Binary_Tree #Binary_Search_Tree
// #2022_06_21_Time_3_ms_(76.33%)_Space_47.3_MB_(52.92%)

import com_github_leetcode.TreeNode;

/*
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
/**
 * 99 - Recover Binary Search Tree\.
 *
 * Medium
 *
 * 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:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/10/28/recover1.jpg)
 *
 * **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:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/10/28/recover2.jpg)
 *
 * **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?
**/
public class Solution {
    private TreeNode prev = null;
    private TreeNode first = null;
    private TreeNode second = null;

    public void recoverTree(TreeNode root) {
        evalSwappedNodes(root);
        int temp = first.val;
        first.val = second.val;
        second.val = temp;
    }

    private void evalSwappedNodes(TreeNode curr) {
        if (curr == null) {
            return;
        }
        evalSwappedNodes(curr.left);
        if (prev != null && prev.val > curr.val) {
            if (first == null) {
                first = prev;
            }
            second = curr;
        }
        prev = curr;
        evalSwappedNodes(curr.right);
    }
}