LeetCode 99. Recover Binary Search Tree Add to List

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?

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public class Solution {    public void recoverTree(TreeNode root) {                if(root == null){            return;        }        //利用中序遍历的上升特性        Stack<TreeNode> ts = new Stack<TreeNode>();        TreeNode ptr = root;        //记录第一次出现乱序的地方        boolean flag = true;        TreeNode sw1 = null, sw2 = null, pre = null;        while( ptr != null || !ts.isEmpty()){            while(ptr != null){                ts.push(ptr);                ptr = ptr.left;            }            if(!ts.isEmpty()){                //栈顶弹栈                ptr = ts.pop();  //当前访问的节点                //第一次出现乱序的地方，用flag来标志第一次乱序的地方，找到后，将flag标为false                if(pre!= null && flag && pre.val > ptr.val){                    //此时错误结点为pre                    sw1 = pre;                    flag = false;                }                //最后一次出现乱序的地方                if(pre != null && pre.val > ptr.val){                    //此时错误结点为当前结点ptr                    sw2 = ptr;  //只要满足前驱大于后继，就记录后继节点，不管是第一次乱序还是第二次乱序                }                pre = ptr;   //pre是ptr的前驱节点                ptr = ptr.right;            }        }        //交换两个节点的值        int val = sw1.val;        sw1.val = sw2.val;        sw2.val = val;    }}


                                             
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