LeetCode OJ 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?

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}".

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先求出树的中序序列，然后在序列中寻找出错的位置。代码如下：

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public void recoverTree(TreeNode root) {
         List<TreeNode> inorder = new ArrayList<TreeNode>();
         inOrder(root, inorder);                     //求中序序列

         TreeNode wrong1 = null;
         TreeNode wrong2 = null;

         for(int i = 0; i < inorder.size() - 1; i++){
             if(inorder.get(i).val > inorder.get(i+1).val){
                 if(wrong1 == null){
                     wrong1 = inorder.get(i);
                     wrong2 = inorder.get(i+1);
                 }
                 else{
                     wrong2 = inorder.get(i+1);
                     break;
                 }
             }
         }
         if(wrong1 != null && wrong2 != null){
             int temp = wrong1.val;
             wrong1.val = wrong2.val;
             wrong2.val = temp;
         }
     }

     public void inOrder(TreeNode root, List<TreeNode> inorder){
         if(root == null) return;
         if(root.left != null) inOrder(root.left, inorder);
         inorder.add(root);
         if(root.right != null) inOrder(root.right, inorder);
     }
 }