题解 | #二叉搜索树与双向链表#

忘记前驱和后继的含义了。。。后面查了才发现自己排反了。

class Solution {
  public:
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if (!pRootOfTree) return pRootOfTree;
        if (!pRootOfTree->left && !pRootOfTree->right) return pRootOfTree;
        TreeNode* head = pRootOfTree;
        TreeNode* lhead = pRootOfTree->left, * rhead = pRootOfTree->right;
        TreeNode* tail;
        if (lhead) {
            head = Convert(lhead);
            for (tail = head; tail->right; tail = tail->right);
            tail->right = pRootOfTree;
            pRootOfTree->left = tail;
        }
        if (rhead) {
            rhead = Convert(rhead);
            rhead->left = pRootOfTree;
            pRootOfTree->right = rhead;
        }
        return head;
    }
};