C 解法, 执行用时: 2ms, 内存消耗: 328KB, 提交时间: 2020-10-28

/**
 * struct TreeNode {
 *  int val;
 *  struct TreeNode *left;
 *  struct TreeNode *right;
 * };
 */
 
/**
 *
 * @param pRootOfTree TreeNode类
 * @return TreeNode类
 */
struct TreeNode* Convert(struct TreeNode* pRootOfTree ) {
    if(!pRootOfTree)
        return pRootOfTree;
    struct TreeNode* p=pRootOfTree;
    if(pRootOfTree->left)
    {
        Convert(pRootOfTree->left);
        p=pRootOfTree->left;
        while(p->right)
            p=p->right;
        pRootOfTree->left=p;
        p->right=pRootOfTree;
    }
    if(pRootOfTree->right)
    {
        Convert(pRootOfTree->right);
        p=pRootOfTree->right;
        while(p->left)
            p=p->left;
        pRootOfTree->right=p;
        p->left=pRootOfTree;
    } 
    while(p->left)
        p=p->left;
    return p;
}

C 解法, 执行用时: 2ms, 内存消耗: 328KB, 提交时间: 2020-08-23

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

/**
 * 
 * @param pRootOfTree TreeNode类 
 * @return TreeNode类
 */
void ConvertList(struct TreeNode *t, struct TreeNode **pre)
{
    if(t == NULL)
        return;
    ConvertList(t->left, pre);
    t->left = *pre;
    if(*pre != NULL)
        (*pre)->right = t;
    *pre = t;
    ConvertList(t->right, pre);
}
struct TreeNode* Convert(struct TreeNode* pRootOfTree ) {
    // write code here
        if(pRootOfTree == NULL)
            return NULL;
    struct TreeNode *pre=NULL;
    ConvertList(pRootOfTree,&pre);
    struct TreeNode* head=pre;
    while(head->left!=NULL){
        head=head->left;
    }
    return head;
}

C 解法, 执行用时: 2ms, 内存消耗: 332KB, 提交时间: 2021-05-04

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */
static struct TreeNode *pre;
void my_DG(struct TreeNode *pRoot)
{
    if( !pRoot ) return ;
    
    my_DG( pRoot->left );
    pRoot->left = pre;
    if(pre) pre->right = pRoot;
    pre = pRoot;
    my_DG( pRoot->right );
}
/**
 * 
 * @param pRootOfTree TreeNode类 
 * @return TreeNode类
 */
struct TreeNode* Convert(struct TreeNode* pRootOfTree ) {
    // write code here
    if( !pRootOfTree ) return NULL;
    
    pre = NULL;
    my_DG(pRootOfTree);
    while(pRootOfTree->left)
        pRootOfTree = pRootOfTree->left;
    return pRootOfTree;
}

C 解法, 执行用时: 2ms, 内存消耗: 336KB, 提交时间: 2020-09-05

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

/**
 * 
 * @param pRootOfTree TreeNode类 
 * @return TreeNode类
 */
void  in_order(struct TreeNode*root,struct TreeNode**new_list){
    if(!root) return;
    else{
        in_order(root->left,new_list);
        root->left=*new_list;
        if(*new_list){
            (*new_list)->right=root;
        }
        (*new_list)=root;
        in_order(root->right,new_list);
    }
}
struct TreeNode* Convert(struct TreeNode* pRootOfTree ) {
    // write code here
    struct TreeNode*new_list=NULL;
    in_order(pRootOfTree,&new_list);
    while(new_list&&new_list->left)
        new_list=new_list->left;
    return new_list;
}

C 解法, 执行用时: 2ms, 内存消耗: 340KB, 提交时间: 2020-11-29

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

/**
 * 
 * @param pRootOfTree TreeNode类 
 * @return TreeNode类
 */
void ConvertList(struct TreeNode *t,struct TreeNode **pre)
{
    if(t==NULL)
        return ;
    ConvertList(t->left,pre);
    t->left=*pre;
    if(*pre!=NULL)
        (*pre)->right=t;
    *pre=t;
    ConvertList(t->right,pre);
}


struct TreeNode* Convert(struct TreeNode* pRootOfTree ) 
{
    // write code here
    if(pRootOfTree==NULL)
        return NULL;
   struct TreeNode *pre=NULL;
    ConvertList(pRootOfTree,&pre);
   struct TreeNode *pHead=pre;
    while(pHead->left!=NULL)
        pHead=pHead->left;
    return pHead;
}