[LeetCode] 99. Recover Binary Search Tree

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

Recover the tree without changing its structure.

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

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    TreeNode lastVisit;
    TreeNode firstNode, secondNode;
    public void recoverTree(TreeNode root) {
        traverse(root);
        int temp = firstNode.val;
        firstNode.val = secondNode.val;
        secondNode.val = temp;
    }
    
    public void traverse(TreeNode root) {
        if (root == null) {
            return;
        }
        traverse(root.left);
        if (lastVisit != null && lastVisit.val > root.val) {
            if (firstNode == null) {
                firstNode = lastVisit;
            } 
            if (firstNode != null) { // not else
                secondNode = root;
            }
        }
        lastVisit = root;
        traverse(root.right);
    }
}

 

/**
 * 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) {
        List<Integer> values = new ArrayList<>();
        List<TreeNode> nodes = new ArrayList<>();
        inorder(root, values, nodes);
        Collections.sort(values);
        for (int i = 0; i < nodes.size(); i++) {
            nodes.get(i).val = values.get(i);
        }
    }
    
    private void inorder(TreeNode root, List<Integer> values, List<TreeNode> nodes) {
        if (root == null) {
            return;
        }
        inorder(root.left, values, nodes);
        values.add(root.val);
        nodes.add(root);
        inorder(root.right, values, nodes);
    }
}