99. Recover Binary Search Tree - Hard

You are given the root of a binary search tree (BST), where exactly two nodes of the tree were swapped by mistake. Recover the tree without changing its structure.

Follow up: A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

 

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

 

inorder traversal, time = O(n), space = O(n)

class Solution {
    public void recoverTree(TreeNode root) {
        List<TreeNode> nodes = new ArrayList<>();
        List<Integer> vals = new ArrayList<>();

        inOrder(root, nodes, vals);
        Collections.sort(vals);

        for(int i = 0; i < nodes.size(); i++) {
            nodes.get(i).val = vals.get(i);
        }
    }

    private void inOrder(TreeNode root, List<TreeNode> nodes, List<Integer> vals) {
        if(root == null) {
            return;
        }
        inOrder(root.left, nodes, vals);
        nodes.add(root);
        vals.add(root.val);
        inOrder(root.right, nodes, vals);
    }
}

 

class Solution {
    TreeNode first = null, second = null, prev = new TreeNode(Integer.MIN_VALUE);

    public void recoverTree(TreeNode root) {
        inOrder(root);

        int tmp = first.val;
        first.val = second.val;
        second.val = tmp;
    }

    private void inOrder(TreeNode root) {
        if(root == null) {
            return;
        }

        inOrder(root.left);

        if(first == null && prev.val >= root.val) {
            first = prev;
        }
        if(first != null && prev.val >= root.val) {
            second = root;
        }
        prev = root;

        inOrder(root.right);
    }
}