897. Increasing Order Search Tree

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.

意思就是把就是按照先序遍历重新构成一个只有右子树的树

思路：

先创建一个空的头节点node，再定义一个指向node的tree（当前节点）然后递归遍历，直到碰到递归边界，就return，然后把当前节点放到tree的右子树上，并让tree重新指向自己的右子树，这样一层一层下来就行了
代码

class Solution:
    tree=TreeNode(None)#头指针
    node=tree#当前节点
    
    def increasingBST(self, root: TreeNode) -> TreeNode:
        if not root:#如果到递归边界
            return
        self.increasingBST(root.left)#遍历左子树
        self.node.right=TreeNode(root.val)#创建节点放到当前
        self.node=self.node.right#移动当前节点
        self.increasingBST(root.right)#向右递归
        return self.tree.right