99. Recover Binary Search Tree

Description

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: [1, 3, null, null, 2]

Output: [3, 1, null, null, 2]

Example 2

Input: [3, 1, 4, null, null, 2]

Output: [2, 1, 4, null, null, 3]

Solutions

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;
    }
}

Solution 1

/**
 * Time complexity : O(1) in the best case, and O(N) in the worst case when 
 *    one of the swapped nodes is a rightmost leaf.
 * Space complexity : up to O(H) to keep the recursion stack, 
 *    where H is a tree height.
 */

class Solution {
    private TreeNode first, last, prev;
    
    public void recoverTree(TreeNode root) {
        if(root == null) return;

        findFaultyNodes(root);
        
        if(first != null && last != null) {
            int val = first.val;
            first.val = last.val;
            last.val = val;
        }
    }
    
    // Basically inorder traversal
    private void findFaultyNodes(TreeNode root) {
        if(root == null) return;
        
        findFaultyNodes(root.left);
        
        if(prev == null) {
            prev = root;
        } else {
            if(prev.val > root.val) {
                if(first == null) {
                    first = prev;
                }
                last = root;
            }
            prev = root;
        }
        
        findFaultyNodes(root.right);
    }
}

Solution 2

/**
 * Time complexity : O(1) in the best case, and O(N) in the worst case when 
 *    one of the swapped nodes is a rightmost leaf.
 * Space complexity : up to O(H) to keep the recursion stack, 
 *    where H is a tree height.
 */
 
 class Solution {  
    public void recoverTree(TreeNode root) {
        if(root == null) return;
        
        TreeNode first = null, last = null, prev = null;
        Deque<TreeNode> stack = new ArrayDeque();
        
        while(!stack.isEmpty() || root != null) {
            while(root != null) {
                stack.add(root);
                root = root.left;
            }
            root = stack.removeLast();

            if(prev != null && prev.val > root.val) {
                last = root;
                if(first == null) {
                    first = prev;
                } else {
                    break;
                }
            }
            
            prev = root;
            root = root.right;
        }
        
        if(first != null && last != null) {
            int val = first.val;
            first.val = last.val;
            last.val = val;
        }
    }
}

Follow up

• A solution using O(n) space is pretty straight forward.

• Could you devise a constant space solution?