题解 | #二叉搜索树与双向链表#

中序遍历,将每个结点放到数组中,然后遍历数组中的元素

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    vector<TreeNode*> vi;
    void dfs(TreeNode* node) {
        if (node) {
            dfs(node -> left);
            vi.push_back(node);
            dfs(node -> right);
        }
    }
    TreeNode* Convert(TreeNode* pRootOfTree) {
        if (!pRootOfTree) return nullptr;
        dfs(pRootOfTree);
        for (int i=0; i<vi.size(); i++) {
//             cout<<vi[i]->val<<endl;
            if (i == 0) {
                vi[i] -> left = nullptr;
            } else {
                vi[i] -> left = vi[i-1];
            }
            if (i == vi.size() - 1) {
                vi[i] -> right = nullptr;
            } else {
                vi[i] -> right = vi[i+1];
            }
        }
        return vi[0];
    }
};