[leetcode]Recover Binary Search Tree

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? 

算法思路：

思路1：求出中序遍历序列，空间复杂度O(n)，tip要求只可以开O(1)空间。

代码略

思路2：中序遍历，遇到每个点，都记录其前驱，记录当前指针cur的前一个节点pre，如果pre.val大于cur.val，表示有错序，多数情况错序有两次；如果有一次错序，说明就是相邻节点需要被交换。

public class Solution {
    TreeNode first = null,second = null,pre = null;
    public void recoverTree(TreeNode root) {
        findNode(root);
        swap(first, second);
    }
    private void findNode(TreeNode root){
        if(root == null){
            return;
        }
        if(root.left != null )  findNode(root.left);
        if(pre != null && root.val < pre.val){
            if(first == null){
                first = pre;
            }
            second = root;
        }
        pre = root;
        if(root.right != null)  findNode(root.right);
    }
    private void swap(TreeNode node1,TreeNode node2){
        int tem = node1.val;
        node1.val = node2.val;
        node2.val = tem;
    }
}