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?

Solution1：In order dfs

思路： 看哪个违反增序，实现时考虑inorder相邻情况
Time Complexity: O(N) Space Complexity: O(N)

Solution2：Morris Traversal

TBC
Time Complexity: O(N) Space Complexity: O(1)

Solution1 Code：

class Solution {
    private TreeNode node1, node2;
    private TreeNode prev = new TreeNode(Integer.MIN_VALUE);
    public void recoverTree(TreeNode root) {
        if(root == null) return;
        
        inorderCheck(root);
        
        int tmp = node1.val;
        node1.val = node2.val;
        node2.val = tmp;
    }
    
    private void inorderCheck(TreeNode node) {
        if(node == null) return;
        
        inorderCheck(node.left);
        
        // in_order
        if(node1 == null && node.val <= prev.val) {
            node2 = node;
            node1 = prev;
        }
        else if(node1 != null && node.val <= prev.val) {
            node2 = node;
        }
        prev = node;
        
        inorderCheck(node.right);
    }
}