[이준열] Day 13

leetcode2/1easy/이준열/94.cpp

@@ -0,0 +1,36 @@
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+ *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
+ * };
+ */
+class Solution {
+public:
+    vector<int> inorderTraversal(TreeNode* root) {
+        vector<int> output;
+        stack<TreeNode*> s;
+        TreeNode *cur = root;
+
+        while (cur || !s.empty())
+        {
+            while (cur)
+            {
+                s.push(cur);
+                cur = cur->left;
+            }
+
+            cur = s.top();
+            s.pop();
+
+            output.push_back(cur->val);
+
+            cur = cur->right;
+        }
+        return output;
+    }
+};

leetcode2/2medium/이준열/200.cpp

@@ -0,0 +1,49 @@
+class Solution {
+public:
+    int dx[4] = {1, 0, -1, 0};
+    int dy[4] = {0, 1, 0, -1};
+    bool visited[300][300] = {};
+
+    void bfs(vector<vector<char>>& grid, int y, int x)
+    {
+        queue<pair<int, int>> q;
+        visited[y][x] = true;
+        q.push(make_pair(y,x));
+
+        while (!q.empty())
+        {
+            pair<int, int> p = q.front();
+            q.pop();
+
+            for (int i = 0; i < 4; i++)
+            {
+                int tmpY = p.first + dy[i];
+                int tmpX = p.second + dx[i];
+                if (tmpY < 0 || tmpY >= grid.size() || tmpX < 0 || tmpX >= grid[0].size())
+                    continue;
+                if (grid[tmpY][tmpX] == '1' && visited[tmpY][tmpX] == false)
+                {
+                    q.push(make_pair(tmpY, tmpX));
+                    visited[tmpY][tmpX] = true;
+                }
+            }
+        }
+    }
+
+
+    int numIslands(vector<vector<char>>& grid) {
+        int output = 0;
+        for (int i = 0; i < grid.size(); i++)
+        {
+            for (int j = 0; j < grid[0].size(); j++)
+            {
+                if (grid[i][j] == '1' && visited[i][j] == false)
+                {
+                    output++;
+                    bfs(grid, i, j);
+                }
+            }
+        }
+        return output;
+    }
+};