recover-binary-search-tree Java code

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}”.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    public void recoverTree(TreeNode root) {
        //可以和上一題 的思路一樣，還是通過中序周遊 來做
        ArrayList<TreeNode> list = new ArrayList<TreeNode>();
        if(root == null){
            return;
        }
        inorder(root,list);
        //找到兩個出錯的節點
        int i;
        int j;
        for ( i = ; i < list.size() - ; i ++) {
            if(list.get(i).val > list.get(i + ).val) {
                break;
            }
        }
        for (j = list.size() - ; j > ; j --) {
            if(list.get(j).val < list.get(j - ).val) {
                break;
            }
        }
        int temp = list.get(i).val;
        list.get(i).val = list.get(j).val;
        list.get(j).val = temp;
    }
    private static void inorder(TreeNode root,List list){
        if(root != null){
            inorder(root.left,list);
            list.add(root);
            inorder(root.right,list);
        }
    }
}