题解 | #二叉搜索树与双向链表#

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution {
  public:
    TreeNode* pre;
    void traversal(TreeNode* root) {
        if (root) {
            traversal(root->left);
            root->left = pre;
            if (pre) {
                pre->right = root;
            }
            pre = root;
            traversal(root->right);
        }
    }
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if (!pRootOfTree) return nullptr;
        TreeNode* p = pRootOfTree;
        while (p->left) {
            p = p->left;
        }
        traversal(pRootOfTree);
        return p;
    }
};