Contents

Basic DFS

Depth-First Search explores as far as possible before backtracking.

// DFS on graph (adjacency list)
void dfs(vector<vector<int>>& graph, int node, vector<bool>& visited) {
    visited[node] = true;
    
    // Process node
    cout << node << " ";
    
    // Explore neighbors
    for (int neighbor : graph[node]) {
        if (!visited[neighbor]) {
            dfs(graph, neighbor, visited);
        }
    }
}

// DFS with return value
bool dfs(vector<vector<int>>& graph, int node, int target, vector<bool>& visited) {
    if (node == target) return true;
    visited[node] = true;
    
    for (int neighbor : graph[node]) {
        if (!visited[neighbor] && dfs(graph, neighbor, target, visited)) {
            return true;
        }
    }
    
    return false;
}

DFS on Grid

DFS for 2D grid problems (connected components, paths).

// DFS on 2D grid (4-directional)
void dfsGrid(vector<vector<char>>& grid, int i, int j) {
    int m = grid.size(), n = grid[0].size();
    
    if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] != '1') {
        return;
    }
    
    grid[i][j] = '0'; // Mark as visited
    
    // Explore 4 directions
    dfsGrid(grid, i + 1, j);
    dfsGrid(grid, i - 1, j);
    dfsGrid(grid, i, j + 1);
    dfsGrid(grid, i, j - 1);
}

// Number of Islands using DFS
int numIslands(vector<vector<char>>& grid) {
    int m = grid.size(), n = grid[0].size();
    int count = 0;
    
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            if (grid[i][j] == '1') {
                count++;
                dfsGrid(grid, i, j);
            }
        }
    }
    
    return count;
}

// Word Search
bool dfsWordSearch(vector<vector<char>>& board, int i, int j, string& word, int idx) {
    if (idx == word.size()) return true;
    if (i < 0 || i >= board.size() || j < 0 || j >= board[0].size()) return false;
    if (board[i][j] != word[idx]) return false;
    
    char temp = board[i][j];
    board[i][j] = '#'; // Mark as visited
    
    vector<pair<int, int>> dirs = \{\{0,1\}, \{0,-1\}, \{1,0\}, \{-1,0\}\};
    for (auto& [dx, dy] : dirs) {
        if (dfsWordSearch(board, i + dx, j + dy, word, idx + 1)) {
            return true;
        }
    }
    
    board[i][j] = temp; // Backtrack
    return false;
}
ID Title Link Solution
200 Number of Islands Link Solution
79 Word Search Link Solution
695 Max Area of Island Link Solution
133 Clone Graph Link Solution
417 Pacific Atlantic Water Flow Link Solution
323 Number of Connected Components Link Solution
547 Number of Provinces Link Solution

DFS on Tree

DFS for tree problems (preorder, inorder, postorder).

// Preorder DFS
void preorder(TreeNode* root, vector<int>& result) {
    if (!root) return;
    result.push_back(root->val);
    preorder(root->left, result);
    preorder(root->right, result);
}

// Inorder DFS
void inorder(TreeNode* root, vector<int>& result) {
    if (!root) return;
    inorder(root->left, result);
    result.push_back(root->val);
    inorder(root->right, result);
}

// Postorder DFS
void postorder(TreeNode* root, vector<int>& result) {
    if (!root) return;
    postorder(root->left, result);
    postorder(root->right, result);
    result.push_back(root->val);
}

// Path Sum
bool hasPathSum(TreeNode* root, int targetSum) {
    if (!root) return false;
    if (!root->left && !root->right) {
        return root->val == targetSum;
    }
    return hasPathSum(root->left, targetSum - root->val) ||
           hasPathSum(root->right, targetSum - root->val);
}

// Sum Root to Leaf Numbers
int sumNumbers(TreeNode* root, int sum = 0) {
    if (!root) return 0;
    sum = sum * 10 + root->val;
    if (!root->left && !root->right) return sum;
    return sumNumbers(root->left, sum) + sumNumbers(root->right, sum);
}
ID Title Link Solution
100 Same Tree Link Solution
101 Symmetric Tree Link Solution
104 Maximum Depth of Binary Tree Link Solution
111 Minimum Depth of Binary Tree Link Solution
112 Path Sum Link Solution
129 Sum Root to Leaf Numbers Link Solution
226 Invert Binary Tree Link Solution
236 Lowest Common Ancestor Link Solution
437 Path Sum III Link Solution
690 Employee Importance Link Solution

DFS with Memoization

DFS with caching to avoid recomputation.

// DFS with memoization (e.g., Longest Increasing Path)
int dfsWithMemo(vector<vector<int>>& matrix, int i, int j, 
                vector<vector<int>>& memo, int prev) {
    int m = matrix.size(), n = matrix[0].size();
    
    if (i < 0 || i >= m || j < 0 || j >= n || matrix[i][j] <= prev) {
        return 0;
    }
    
    if (memo[i][j] != -1) {
        return memo[i][j];
    }
    
    int result = 1;
    vector<pair<int, int>> dirs = \{\{0,1\}, \{0,-1\}, \{1,0\}, \{-1,0\}\};
    
    for (auto& [dx, dy] : dirs) {
        result = max(result, 1 + dfsWithMemo(matrix, i + dx, j + dy, 
                                              memo, matrix[i][j]));
    }
    
    memo[i][j] = result;
    return result;
}
ID Title Link Solution
329 Longest Increasing Path in a Matrix Link -

Iterative DFS

DFS using stack instead of recursion.

// Iterative DFS on graph
void dfsIterative(vector<vector<int>>& graph, int start) {
    stack<int> st;
    vector<bool> visited(graph.size(), false);
    
    st.push(start);
    
    while (!st.empty()) {
        int node = st.top();
        st.pop();
        
        if (visited[node]) continue;
        visited[node] = true;
        
        // Process node
        cout << node << " ";
        
        // Push neighbors in reverse order to maintain order
        for (int i = graph[node].size() - 1; i >= 0; --i) {
            if (!visited[graph[node][i]]) {
                st.push(graph[node][i]);
            }
        }
    }
}

// Iterative DFS on tree
vector<int> preorderIterative(TreeNode* root) {
    vector<int> result;
    if (!root) return result;
    
    stack<TreeNode*> st;
    st.push(root);
    
    while (!st.empty()) {
        TreeNode* node = st.top();
        st.pop();
        result.push_back(node->val);
        
        if (node->right) st.push(node->right);
        if (node->left) st.push(node->left);
    }
    
    return result;
}
ID Title Link Solution
144 Binary Tree Preorder Traversal Link -
94 Binary Tree Inorder Traversal Link -