面试题36. 二叉搜索树与双向链表

题目：

 

 

 

 

解答：

// 中序遍历即可。只需要记录一个pre指针即可。

 
class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == nullptr) 
        {
            return nullptr;
        }
        
        TreeNode* pre = nullptr;
         
        convertHelper(pRootOfTree, pre);
         
        TreeNode* res = pRootOfTree;
        while(res ->left)
        {
            res = res ->left;
        }
        return res;
    }
     
    void convertHelper(TreeNode* cur, TreeNode*& pre)
    {
        if(cur == nullptr) 
        {
            return;
        }
         
        convertHelper(cur ->left, pre);
         
        cur ->left = pre;
        if(pre) 
        {
            pre ->right = cur;
        }
        pre = cur;
         
        convertHelper(cur ->right, pre);    
    }
};