二叉树的下一个结点

 1 class Solution {
 2 public:
 3     TreeLinkNode* GetNext(TreeLinkNode* pNode)
 4     {
 5        if(pNode == NULL )
 6            return NULL;
 7         TreeLinkNode *p= NULL;
 8         if(pNode->right == NULL)
 9             { 
10                p = pNode;
11                while(p->next && p != p->next->left)
12                     {
13                       p = p->next;
14                     }
15               
16                 return p->next;
17             }
18         else 
19         {
20             p = pNode->right;
21             while(p->left != NULL)
22                   p = p->left;
23         }
24           return p;
25     }
26 };
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原文地址：https://www.cnblogs.com/daocaorenblog/p/5463274.html