思路:
按照规律,即为二叉树中序遍历的逆序,从右子树开始,一次把所有的节点累加,加到当前节点
class Solution(object):
def convertBST(self, root):
self.num = 0
if not root:
return None
def DFS(root):
if not root:
return
DFS(root.right)
root.val += self.num
self.num = root.val
DFS(root.left)
DFS(root)
return root