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?

想法很简单，找到两个顺序不对的点，最后交换。但是如何知道这两个点的顺序是不对的呢？因为BST是左<中<右，所以我们可以用in-order traverse(中序遍历)的方法：
其中In-order traverse模板：

public void inOrderTraverse{
    inOrderTraverse(root.left);
    //do something
    inOrderTraverse(root.right);
}

维护一个preNode节点，这个节点代表当前访问root节点的中序遍历（就是顺序排序）相对左边的一个，比如结果是45678的二叉树，root =7的时候，preNode就是6，root是8的时候，preNode=7。

Inorder traveral will return values in an increasing order. So if an element is less than its previous element,the previous element is a swapped node OR this element is a swapped node.

被交换的两个节点分别设为firstNode,secondNode,firstNode代表的是相对于secondNode偏左的节点，二者是被调换位置了。所以firstNode是相对root左边的节点preNode，secondNode是相对preNode右边的节点root。

Runtime: 4 ms  beats 41.66% of java submissions.

	static TreeNode firstNode = null;
	static TreeNode secondNode = null;
	static TreeNode preNode = new TreeNode(Integer.MIN_VALUE);

	public static void recoverTree(TreeNode root) {
		inorder(root);
		int tmp = firstNode.val;
		firstNode.val = secondNode.val;
		secondNode.val = tmp;
	}

	public static void inorder(TreeNode root) {
		if (root == null) return;
		inorder(root.left);
		if (root.val <= preNode.val) {
			if (firstNode == null) firstNode = preNode;
			secondNode = root;
		}
		preNode = root;
		inorder(root.right);
	}