LeetCode Flatten Binary Tree to Linked List

//  56ms  如果不加  root->left=NULL;  则出现runtime error  应该是结束之后 这样的树有左节点  所以 程序比较两个树是否相同时用了 NULL->left

 1 /**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    
    void ReFlat(TreeNode *root,TreeNode **head,TreeNode **end)
    {
        TreeNode *leftHead,*leftEnd,*rightHead,*rightEnd;
        *head=root;
        *end=root;
        leftHead=leftEnd=rightHead=rightEnd=NULL;
        if(root->left)
        {
            ReFlat(root->left,&leftHead,&leftEnd);       
        }  
        if(root->right)
        {
            ReFlat(root->right,&rightHead,&rightEnd);        
        }
        if(leftHead)
        {
            (*end)->right=leftHead;
            *end=leftEnd;
        }
        if(rightHead)
        {
            (*end)->right=rightHead;
            (*end)=rightEnd;
        }
        root->left=NULL;
    }
    void flatten(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        TreeNode *head,*end;
        head=end=NULL;
        if(root==NULL)
            return;
        ReFlat(root,&head,&end);
        
        
        
    }
};