Recover Binary Search Tree 07/26

In a binary search tree, (Only) two nodes are swapped. Find out these nodes and swap them. If there no node swapped, return original root of tree.Have you met this question in a real interview? Yes

Example

Given a binary search tree:

    4
   / \
  5   2
 / \
1   3

return

    4
   / \
  2   5
 / \
1   3

Based on inorder traversal. O(N)

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root: the given tree
     * @return: the tree after swapping
     */
     
     public TreeNode first = null;
    public TreeNode second = null;
    public TreeNode last = new TreeNode(Integer.MIN_VALUE);
   
   
   
    public TreeNode bstSwappedNode(TreeNode root) {
        // write your code here
        if(root == null){
            
            return null;
        }
        
        traversal(root);
        
        if(first != null && second != null){
            int tmp = first.val;
            first.val = second.val;
            second.val = tmp;
        }
        
        
        return root;
       
    }
    
    
    
    public void traversal(TreeNode root){
        if(root == null){
            
            return;
        }
        
        traversal(root.left);
        
        if(first == null && root.val < last.val){
            
            first = last;
        }
        if(first != null && root.val < last.val){
            second = root;
        }
        
        last = root;
        traversal(root.right);
        
    }
}