Depth-First Search (DFS) is one of the most fundamental graph traversal algorithms. It works by starting at a node and exploring as far down each branch as possible before backtracking — making it ideal for problems involving reachability, connected components, paths, and tree structure. This page collects ready-to-use C++ templates for the most common DFS patterns you’ll encounter on LeetCode. See also Graph and Backtracking.

New to DFS? DFS explores as deep as possible before backtracking. Think of it like exploring a maze — go straight until you hit a dead end, then back up and try the next turn.

DFS Traversal Order 1 ① 2 ② 3 ⑤ 4 ③ 5 ④ 6 ⑥ Stack 3 5 ← top of stack Visited Processing Unvisited DFS goes deep before going wide

Contents

Basic DFS

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

When to use: Checking reachability between nodes, finding connected components, or exploring all paths in a general graph.

# DFS on graph (adjacency list)
def dfs_graph(graph: list[list[int]], node: int, visited: list[bool]) -> None:
    visited[node] = True
    # process node
    for neighbor in graph[node]:
        if not visited[neighbor]:
            dfs_graph(graph, neighbor, visited)


def dfs_find_target(
    graph: list[list[int]], node: int, target: int, visited: list[bool]
) -> bool:
    if node == target:
        return True
    visited[node] = True
    for neighbor in graph[node]:
        if not visited[neighbor] and dfs_find_target(graph, neighbor, target, visited):
            return True
    return False
ID Title Link Solution
841 Keys and Rooms Link Solution

DFS on Grid

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

When to use: Flood-fill problems, island counting, or any task where you explore connected cells in a 2D matrix.

DFS flood fill on a grid 0 1 2 3 4 1 ① 1 ② 0 0 0 0 1 ③ 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 DFS flood fill Start at (0,0), explore all connected 1s: ① (0,0) → go right ② (0,1) → go down ③ (1,1) → no more 1s Backtrack → island done! Explored (flood fill) Land (not yet visited) Water (0) Grid has 2 islands. DFS marks each connected component of 1s. 
# DFS on 2D grid (4-directional)
DIRS = [(0, 1), (0, -1), (1, 0), (-1, 0)]


def dfs_grid(grid: list[list[str]], i: int, j: int) -> None:
    m, n = len(grid), len(grid[0])
    if i < 0 or i >= m or j < 0 or j >= n or grid[i][j] != "1":
        return
    grid[i][j] = "0"
    for di, dj in DIRS:
        dfs_grid(grid, i + di, j + dj)


def num_islands(grid: list[list[str]]) -> int:
    m, n = len(grid), len(grid[0])
    count = 0
    for i in range(m):
        for j in range(n):
            if grid[i][j] == "1":
                count += 1
                dfs_grid(grid, i, j)
    return count


def dfs_word_search(board: list[list[str]], i: int, j: int, word: str, idx: int) -> bool:
    if idx == len(word):
        return True
    if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]):
        return False
    if board[i][j] != word[idx]:
        return False
    temp = board[i][j]
    board[i][j] = "#"
    for di, dj in DIRS:
        if dfs_word_search(board, i + di, j + dj, word, idx + 1):
            board[i][j] = temp
            return True
    board[i][j] = temp
    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).

When to use: Tree traversals, path-sum problems, computing tree height/diameter, or any recursive tree decomposition.

class TreeNode:
    def __init__(self, val: int = 0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


def preorder(root: TreeNode | None, result: list[int]) -> None:
    if not root:
        return
    result.append(root.val)
    preorder(root.left, result)
    preorder(root.right, result)


def inorder(root: TreeNode | None, result: list[int]) -> None:
    if not root:
        return
    inorder(root.left, result)
    result.append(root.val)
    inorder(root.right, result)


def postorder(root: TreeNode | None, result: list[int]) -> None:
    if not root:
        return
    postorder(root.left, result)
    postorder(root.right, result)
    result.append(root.val)


def has_path_sum(root: TreeNode | None, target_sum: int) -> bool:
    if not root:
        return False
    if not root.left and not root.right:
        return root.val == target_sum
    return has_path_sum(root.left, target_sum - root.val) or has_path_sum(
        root.right, target_sum - root.val
    )


def sum_numbers(root: TreeNode | None, cur: int = 0) -> int:
    if not root:
        return 0
    cur = cur * 10 + root.val
    if not root.left and not root.right:
        return cur
    return sum_numbers(root.left, cur) + sum_numbers(root.right, cur)
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.

When to use: Problems with overlapping subproblems on graphs or grids, such as longest increasing path or counting distinct paths.

DFS with memoization — Longest Increasing Path col 0 col 1 col 2 9 memo=1 9 memo=1 4 memo=2 6 memo=2 6 memo=2 8 memo=1 2 memo=3 1 memo=4 ★ 1 memo=2 Longest increasing path 1 → 2 → 6 → 9 (length 4, shown with arrows) How memo works: memo[r][c] = length of the longest path starting from (r,c) DFS from cell 1 at (2,1): → can go to 2: 1+memo[2][0] → memo[2][0]=3 (2→6→9) → so memo[2][1] = 1+3 = 4 On longest path Path start (max memo) 
DIRS4 = [(0, 1), (0, -1), (1, 0), (-1, 0)]


def dfs_with_memo(
    matrix: list[list[int]], i: int, j: int, memo: list[list[int]], prev: int
) -> int:
    m, n = len(matrix), len(matrix[0])
    if i < 0 or i >= m or j < 0 or j >= n or matrix[i][j] <= prev:
        return 0
    if memo[i][j] != -1:
        return memo[i][j]
    best = 1
    for di, dj in DIRS4:
        best = max(best, 1 + dfs_with_memo(matrix, i + di, j + dj, memo, matrix[i][j]))
    memo[i][j] = best
    return best
ID Title Link Solution
329 Longest Increasing Path in a Matrix Link Solution

Iterative DFS

DFS using stack instead of recursion.

When to use: When the recursion depth might cause a stack overflow, or when you need explicit control over the traversal order.

class TreeNode:
    def __init__(self, val: int = 0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


def dfs_iterative(graph: list[list[int]], start: int) -> None:
    st: list[int] = [start]
    visited = [False] * len(graph)
    while st:
        node = st.pop()
        if visited[node]:
            continue
        visited[node] = True
        for nei in reversed(graph[node]):
            if not visited[nei]:
                st.append(nei)


def preorder_iterative(root: TreeNode | None) -> list[int]:
    if not root:
        return []
    result: list[int] = []
    st: list[TreeNode] = [root]
    while st:
        node = st.pop()
        result.append(node.val)
        if node.right:
            st.append(node.right)
        if node.left:
            st.append(node.left)
    return result
ID Title Link Solution
144 Binary Tree Preorder Traversal Link -
94 Binary Tree Inorder Traversal Link -

Pattern Comparison

Pattern When to Use Time Space
Basic DFS Reachability, connected components O(V+E) O(V)
Grid DFS Flood fill, island counting O(M×N) O(M×N)
Tree DFS All tree traversals, path problems O(N) O(H)
DFS + Memo Overlapping subproblems on graphs/grids O(States) O(States)
Iterative When recursion stack overflows O(V+E) O(V)

DFS vs BFS

When should you pick DFS over BFS (or vice versa)?

  • Use DFS when you need to explore all paths, check connectivity, detect cycles, or solve problems that decompose recursively (e.g., tree shape, backtracking). DFS is also more memory-efficient on narrow/deep structures.
  • Use BFS when you need the shortest path in an unweighted graph, want to process nodes level by level, or need the minimum number of steps to reach a target.
  • Rule of thumb: If the problem says “shortest” or “minimum steps,” reach for BFS. If it says “all paths,” “connected,” or “exists,” DFS is usually the natural fit.
DFS — Goes deep first A B C D E F ① ② ③ ④ ⑤ ⑥ deep vs BFS — Goes wide first level 1 A B C D E F ① ② ③ ④ ⑤ ⑥ wide

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