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给定一个树，

按中序遍历

重新排列树，使树中最左边的结点现在是树的根，并且每个结点没有左子结点，只有一个右子结点。

示例 ：

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

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

输出：[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  

提示：

1. 给定树中的结点数介于 1 和 100 之间。
2. 每个结点都有一个从 0 到 1000 范围内的唯一整数值。

中序遍历，然后新建一棵树

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def increasingBST(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        mid = []
        def midOrd(root):
            if root:
                midOrd(root.left)
                mid.append(root.val)
                midOrd(root.right)
        midOrd(root)
        
        ans = TreeNode(0)
        p = ans
        for i in mid:
            p.right = TreeNode(i)
            p = p.right
        return ans.right

借助栈模拟中序遍历

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def increasingBST(self, root: TreeNode) -> TreeNode:
        s = []
        ans = TreeNode(0)
        p = ans
        
        while s or root:
            if root:
                s.append(root)
                root = root.left
            else:
                cur = s.pop()
                root = cur.right
                cur.left = None
                p.right = cur
                p = p.right
        return ans.right

还有这种神仙递归

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def fun(self, root, tail = None):
        if not root:
            return tail
        ans = self.fun(root.left, root)
        root.left = None
        root.right = self.fun(root.right, tail)
        return ans
    
    def increasingBST(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        return self.fun(root, None)