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}".

Subscribe to see which companies asked this question

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);
    }
}