题解 | #二叉树的下一个结点#

# -*- 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
		# 如果有右子树，则后继为右子树最左侧的点
        if pNode.right:
            pNode = pNode.right
            while pNode.left: pNode = pNode.left
            return pNode
        # 没有右子树，后继往上找，找到的第一个节点i，满足i是i的father的左孩子
        # i的father就是pNode的后一节点
        while pNode.next and pNode.next.left != pNode: pNode = pNode.next
        return pNode.next