[pull] master from youngyangyang04:master

problems/0063.不同路径II.md

@@ -145,7 +145,7 @@ public:
     int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
         int m = obstacleGrid.size();
         int n = obstacleGrid[0].size();
-	if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) //如果在起点或终点出现了障碍，直接返回0
+        if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) //如果在起点或终点出现了障碍，直接返回0
             return 0;
         vector<vector<int>> dp(m, vector<int>(n, 0));
         for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) dp[i][0] = 1;

problems/0111.二叉树的最小深度.md

@@ -64,7 +64,7 @@
 
 代码如下：
 
-```
+```CPP
 int getDepth(TreeNode* node)
 ```
 
@@ -74,14 +74,14 @@ int getDepth(TreeNode* node)
 
 代码如下：
 
-```
+```CPP
 if (node == NULL) return 0;
 ```
 
 3. 确定单层递归的逻辑
 
 这块和求最大深度可就不一样了，一些同学可能会写如下代码：
-```
+```CPP
 int leftDepth = getDepth(node->left);
 int rightDepth = getDepth(node->right);
 int result = 1 + min(leftDepth, rightDepth);