Given a tree, rearrange the tree in in-order so that the leftmost node in the tree is now the root of the tree, and every node has no left child and only 1 right child.

Example 1:
Input: [5,3,6,2,4,null,8,1,null,null,null,7,9]

       5
      / \
    3    6
   / \    \
  2   4    8
 /        / \ 
1        7   9

Output: [1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

 1
  \
   2
    \
     3
      \
       4
        \
         5
          \
           6
            \
             7
              \
               8
                \
                 9  

Note:

1. The number of nodes in the given tree will be between 1 and 100.
2. Each node will have a unique integer value from 0 to 1000.

M1: 先inorder traverse，再建一個新樹

time: O(n), space: O(N)　　-- N: size of the answer

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 
 
*/
class Solution {
    public TreeNode increasingBST(TreeNode root) {
        if(root == null) {
            return null;
        }
        List<Integer> inorder = inorder(root, new ArrayList<>());
        
        TreeNode dummy = new TreeNode(0);
        TreeNode cur = dummy;
        for
 
(int v : inorder) {
            cur.right = new TreeNode(v);
            cur = cur.right;
        }
        return dummy.right;
    }
    
    public List<Integer> inorder(TreeNode root, List<Integer> list) {
        if(root == null) {
            return null;
        }
        inorder(root.left, list);
        list.add(root.val);
        inorder(root.right, list);
        return list;
    }
}

M2: inorder的同時relink

time: O(n), space: O(height)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    TreeNode cur;
    
    public TreeNode increasingBST(TreeNode root) {
        if(root == null) {
            return null;
        }
        TreeNode dummy = new TreeNode(0);
        cur = dummy;
        inorder(root);
        return dummy.right;
    }
    
    public void inorder(TreeNode node) {
        if(node == null) {
            return;
        }
        inorder(node.left);
        node.left = null;
        cur.right = node;
        cur = node;
        inorder(node.right);
    }
}