99. Recover Binary Search Tree

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

Recover the tree without changing its structure.

Example 1:

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

   1
  /
 3
  \
   2


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

   3
  /
 1
  \
   2

Example 2:

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

  3
 / \
1   4
   /
  2


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

  2
 / \
1   4
   /
  3

Follow up:

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

• Could you devise a constant space solution?

Thoughts:

1. In-order traversal + first, sec, prev node global record

2. Morris traversal to save space

Code 1 T: O(n); S: O(n) due to call stack space:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    TreeNode first;
    TreeNode sec;
    TreeNode prev;
    public void recoverTree(TreeNode root) {
        if(root == null) return;
        inorder_traversal(root);
        // swap the first and sec value
        int tmp = first.val;
        first.val = sec.val;
        sec.val = tmp;

        return;
    }
    private void inorder_traversal(TreeNode cur){
        if(cur == null) return;
        inorder_traversal(cur.left);

        if(first == null && (prev != null && prev.val > cur.val)) first = prev;
        if(first != null && prev.val > cur.val) sec = cur;

        prev = cur;
        inorder_traversal(cur.right);

    }
}

Code 2: T:O(n); S: O(1)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public void recoverTree(TreeNode root) {
        TreeNode prev = null; // previous node in in-order traversal
        TreeNode first = null;
        TreeNode sec = null;

        // Morris Traversal:
        TreeNode pred = null; // previous node in Morris Traversal
        while(root != null){
            // constructing threading
            if (root.left != null){
                 pred = root.left;
                    while(pred.right != null && pred.right != root){
                        pred = pred.right;
                    }

                    if(pred.right == null){
                        // connecting 
                        pred.right = root;
                        root = root.left;
                    }
                    else{
                        // pred.right == root: visited
                        // record the first and second node
                        if(prev.val > root.val){
                            if(first == null) {first = prev; sec = root;}
                            else {sec = root;}
                        }


                        // disconnecting
                        pred.right = null;
                        // updating prev
                        prev = root;
                        root = root.right;

                    }
                }
            else{

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

                prev = root;
                root = root.right;
            }
        }
        // swap first and sec node values
        // if(first != null && sec != null){
            int tmp = first.val;
            first.val = sec.val;
            sec.val = tmp;
        // }

    }

}