题解 | #数组中重复的数字#

暴力解

直接用pNode找到根节点
进行中序遍历
在返回的列表中找到对应的下一个值

# class TreeLinkNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None
class Solution:
    def inorder(self,root,order):
        if root:
            self.inorder(root.left,order)
            order.append(root)
            self.inorder(root.right,order)
            
    def GetNext(self, pNode):
        order=[]
        root=pNode
        while root.next:
            root=root.next
        self.inorder(root,order)
        for i in range(len(order)-1):
            if pNode==order[i]:
                return None if i==len(order)-1 else order[i+1]
            
        # write code here