# 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。
class TreeLinkNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
        self.next = None
class Solution:
    def GetNext(self, pNode):
        temp = None
        # 根节点的情况
        if(pNode.next == None):
            if (pNode.right != None):
                temp = pNode.right
                while temp.left != None:
                    temp = temp.left
                return temp
            else:
                return None
        # 是父节点的左孩子
        if pNode == pNode.next.left:
            temp = None

            if pNode.right != None:
                temp = pNode.right
                while temp.left != None:
                    temp = temp.left
                return temp
            else:
                return pNode.next
        # 是父节点的右孩子
        else:
            temp = None
            #  该节点还有右子树
            if pNode.right != None:
                temp = pNode.right
                while temp.left != None:
                    temp = temp.left
                return temp
            # 该节点没有右子树（返回与该节点最近的一个拥有左孩子的父节点）
            else:
                last = pNode
                temp = pNode.next
                while temp.next != None:
                    temp = temp.next
                    last = last.next
                    if temp.left == last:
                        return temp
                # 上面的所用情况都不满足，即该节点为最后一个，返回None
                return None
a1 = TreeLinkNode(10)
a2 = TreeLinkNode(5)
a3 = TreeLinkNode(12)
a4 = TreeLinkNode(4)
a5 = TreeLinkNode(7)
a1.left = a2
a1.right = a3
a3.left = a4
a3.right = a5
a1.next = None
a2.next = a1
a3.next = a1
a4.next = a3
a5.next = a3
S = Solution()
print S.GetNext(a2).val