[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}".

中序遍历，用一个List存储每个节点的位置，然后双指针一个从左往右，一个从右往左找到连个错误的位置，然后交换连个节点的值。

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