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.

# Definition for a binary tree node.class TreeNode:    def __init__(self, x):        self.val = x        self.left = None        self.right = Noneclass Solution:    def increasingBST(self, root):        """        :type root: TreeNode        :rtype: TreeNode        """        l = []        if root is None:            return None        def inorder(root):            if root==None:                return            if root.left:                inorder(root.left)            l.append(root.val)            if root.right:                inorder(root.right)        inorder(root = root)        newtree = TreeNode(l[0])        build = newtree        for i in l[1:]:            build.right = TreeNode(i)            build = build.right        return newtree