diff --git a/README.md b/README.md
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@@ -171,6 +171,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 208 |[208. Implement Trie (Prefix Tree)](https://leetcode.com/problems/implement-trie-prefix-tree/)|[C++](solutions/208.%20Implement%20Trie%20(Prefix%20Tree).md)|Medium|  |
 | 209 |[Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/209.%20Minimum%20Size%20Subarray%20Sum.md)|Medium|  |
 | 210 |[Course Schedule II](https://leetcode.com/problems/course-schedule-ii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/210.%20Course%20Schedule%20II.md)|Medium|  |
+| 212 |[Word Search II](https://leetcode.com/problems/word-search-ii/)|[C++](solutions/212.%20Word%20Search%20II.md)|Hard|  |
 | 213 |[House Robber II](https://leetcode.com/problems/house-robber-ii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/213.%20House%20Robber%20II.md)|Medium|  |
 | 215 |[Kth Largest Element in an Array](https://leetcode.com/problems/kth-largest-element-in-an-array/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/215.%20Kth%20Largest%20Element%20in%20an%20Array.md)|Medium|  |
 | 216 |[Combination Sum III](https://leetcode.com/problems/combination-sum-iii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/216.%20Combination%20Sum%20III.md)|Medium|  |
diff --git a/solutions/212. Word Search II.md b/solutions/212. Word Search II.md
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@@ -0,0 +1,89 @@
+# [212. Word Search II](https://leetcode.com/problems/word-search-ii/)
+
+# 思路
+
+这题是[79. Word Search](https://leetcode.com/problems/word-search/)的升级版，79题是判断某个单词是否存在与board中，而此题是给了一堆单词，让返回所有存在的单词。
+
+亲测按照79那样直接搜索是会超时的（不然就不是hard题了），那我们只好考虑更加高效的方法。由于是在一个单词集合中查询，那么我们考虑用字典树（Trie），所以我们可以直接把[208题](208.%20Implement%20Trie%20(Prefix%20Tree).md)的字典树实现拿过来用，不过只用到了`insert`操作。
+
+字典树建好后，我们就可以DFS了，一旦遇到了字典树某个节点`is_last_letter == true`就说明找到了，将此时对应的单词加入结果数组中；为了避免下一次DFS时重复找到这个单词，我们应该将这里的`is_last_letter`设为false，表示单词集里面没有这个单词了。
+
+代码中使用了一个小技巧，类似[79题思路二](79.%20Word%20Search.md)那样，我们使用DFS时不另开visited数组，而是用原数组board来记录某个元素是否被访问过，若被访问过就将board对应位置改成字符`*`，注意DFS结束前要改回来。
+
+# C++
+``` C++
+struct TrieNode{
+    bool is_last_letter;    // 是否是某个单词的结尾
+    char letter;
+    vector<TrieNode *>next; // 孩子们
+    TrieNode(char cc): is_last_letter(false), letter(cc),
+                       next(vector<TrieNode *>(26, NULL)){}
+};
+
+class Trie {
+private:
+    TrieNode *root;
+public:
+    Trie() {
+        root = new TrieNode('\0'); // 根节点不包含字符
+        root -> is_last_letter = true;
+    }
+    
+    TrieNode* getRoot(){return root;}
+    
+    /** Inserts a word into the trie. */
+    void insert(string word) {
+        if(word.empty()) return;
+        TrieNode *p = root;
+        for(int i = 0; i < word.size(); i++){
+            int idx = word[i] - 'a';
+            if(NULL == (p -> next)[idx]) // 没找到, 插入即可
+                (p -> next)[idx] = new TrieNode(word[i]);
+            p = (p -> next)[idx];
+        }
+        p -> is_last_letter = true;
+    }
+};
+
+class Solution {
+private:
+    int row, col;
+    vector<string>res;
+    const vector<int>dx = {0,0,1,-1}, dy = {1,-1,0,0};
+    void DFS(vector<vector<char>>& board,
+             vector<TrieNode *>&nodes, int i, int j, string word){
+        if(i < 0 || i >= row || j < 0 || j >= col || board[i][j] == '*') return;
+        
+        TrieNode *node = nodes[board[i][j] - 'a'];
+        if(!node) return;
+        
+        word.append(1, board[i][j]);
+        char tmp = board[i][j]; board[i][j] = '*';
+        if(node -> is_last_letter) {
+            res.push_back(word);
+            // 避免下一次DFS时重复找到这个单词
+            node -> is_last_letter = false;
+        }
+        for(int k = 0; k < 4; k++)
+            DFS(board, node -> next, i+dx[k], j+dy[k], word);
+
+        board[i][j] = tmp;
+    }
+    
+public:
+    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
+        row = board.size();
+        if(!row || words.empty()) return res;
+        col = board[0].size();
+        
+        Trie tree;
+        for(string word: words) tree.insert(word); 
+        
+        for(int i = 0; i < row; i++)
+            for(int j = 0; j < col; j++)
+                DFS(board, tree.getRoot() -> next, i, j, "");
+        
+        return res;
+    }
+};
+```
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