题解 | NO.30#二叉搜索树与双向链表#3.11

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

/**
 * 
 * @param pRootOfTree TreeNode类 
 * @return TreeNode类
 */
//递归中序遍历
static struct TreeNode* pre = NULL;		//静态全局变量
static struct TreeNode* head = NULL;
struct TreeNode* Convert(struct TreeNode* pRootOfTree ) {
    if(pRootOfTree == NULL)
        return NULL;
    Convert(pRootOfTree->left);
    if(pre == NULL){
        head = pRootOfTree;
        pre = pRootOfTree;
    }else{		//将当前节点和前驱结点连接
        pre->right = pRootOfTree;
        pRootOfTree->left = pre;
        pre = pRootOfTree;	//更新前驱节点
    }
    Convert(pRootOfTree->right);
    return head;
}

//Q：为什么要用静态全局变量？单纯的全局变量通不过编译，疑惑