[LeetCode]Recover Binary Search Tree

[LeetCode]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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

中序遍歷，用一個List存儲每個節點的位置，然後雙指針一個從左往右，一個從右往左找到連個錯誤的位置，然後交換連個節點的值。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
	List list = new ArrayList<>();
    public void recoverTree(TreeNode root) {
    	mid(root);
    	int left = -1;
    	int right = -1;
    	for(int i=0;ilist.get(i+1).val){
    			left = i;
    			break;
    		}
    	}
    	for(int j=list.size()-1;j>0;j--){
    		if(list.get(j).val