二叉树的下一个结点.py

'''
题目描述
给定一个二叉树和其中的一个结点，请找出中序遍历顺序的下一个结点并且返回。注意，树中的结点不仅包含左右子结点，同时包含指向父结点的指针。
'''

# -*- coding:utf-8 -*-
# class TreeLinkNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None
class Solution:
    def GetNext(self, pNode):
        # write code here
        if pNode is None:
            return
        pNext = None
        if pNode.right:
            pNode = pNode.right
            while pNode.left:
                pNode = pNode.left
            pNext = pNode
        else:
            if pNode.next and pNode.next.left == pNode:#.next代表指向root节点
                pNext = pNode.next
            elif pNode.next and pNode.next.right == pNode:
                pNode = pNode.next
                while pNode.next and pNode.next.right == pNode:
                    pNode = pNode.next
                if pNode.next:
                    pNext = pNode.next
        return pNext