99. Recover Binary Search Tree

99. Recover Binary Search Tree

https://leetcode.com/problems/recover-binary-search-tree/discuss/32535/No-Fancy-Algorithm-just-Simple-and-Powerful-In-Order-Traversal


https://www.youtube.com/watch?v=LR3K5XAWV5k




/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */



class Solution{
  TreeNode firstElement = null;
  TreeNode secondElement = null;
  
// The reason for this initialization is to avoid null pointer exception in the first comparison when prevElement has not been initialized

  TreeNode prevElement = new TreeNode(Integer.MIN_VALUE);
  
  public void recoverTree(TreeNode root){
    traverse(root);
     
    // after done traversing, swap the first node value and the second node value 
    int tmp = firstElement.val;
    firstElement.val = secondElement.val;
    secondElement.val = tmp;
  }
  
  private void traverse(TreeNode root){
    if(root == null) return;
    
    traverse(root.left);
    
    if(firstElement == null && prevElement.val >= root.val){
      firstElement = prevElement;
    }
    if(firstElement != null && prevElement.val >= root.val){
      secondElement = root;
    }
    
    prevElement = root;
    
    traverse(root.right);
  }
}


// two cases 
// one is if the original bst tree in order is 1 2 3 4 5 6 7 
// after swap : we have 1 6 3 4 5 2 7 
// 6 and 2 are swapped. 
// there are two abonormal looking pairs : which are 6 3 and 5 2 


// anothe case is 
// the original bst in order is 1 2 3 4 5 6 7
// after swap : 1 3 2 4 5 6 7
// 3 and 2 are swapped 
// there is only one pair of abonormal looking pair which is 3 and 2 


// combine above two cases, after we meet the first pair, we don't know 
// if there is only one pair, or there is another pair , so we store the root of the first pair as 
// the second element . just in case there is only one pair
// if there is another pair later, we can just replace the second element with the root of the second pair 

 

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?