How do you solve Binary Tree Paths on LeetCode?

Binary Tree Paths (LeetCode #257) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding tree traversal is crucial for solving tree-related problems.

What is the brute force solution for Binary Tree Paths?

The brute force approach for Binary Tree Paths is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves traversing the entire tree and storing all paths from the root to each leaf node. This is straightforward but inefficient as it may require multiple traversals. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Binary Tree Paths?

Binary Tree Paths uses the following concepts: String, Backtracking, Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Backtracking.

What are the interview tips for Binary Tree Paths?

Clarify the definition of a leaf node before starting. Discuss your thought process and approach before coding. Test your code with different tree structures to ensure correctness.

What are common mistakes when solving Binary Tree Paths?

Not handling leaf nodes correctly. Forgetting to backtrack properly in recursive functions.

#257

Binary Tree Paths

Easy

StringBacktrackingTreeDepth-First SearchBinary TreeDepth-First SearchBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution still uses depth-first search but avoids unnecessary string concatenations by using a list to build the path. This reduces the overhead of string operations and improves efficiency.

⚙️

Algorithm

4 steps

1Step 1: Start from the root and initialize an empty list to store paths.
2Step 2: Use a recursive function to explore each node, appending the current node's value to a list.
3Step 3: If a leaf node is reached, convert the list to a string and add it to the paths list.
4Step 4: Backtrack to explore other branches of the tree.

solution.py21 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8class Solution:
9    def binaryTreePaths(self, root: TreeNode) -> List[str]:
10        paths = []
11        def dfs(node, path):
12            if node:
13                path.append(str(node.val))
14                if not node.left and not node.right:
15                    paths.append('->'.join(path))
16                else:
17                    dfs(node.left, path)
18                    dfs(node.right, path)
19                path.pop()
20        dfs(root, [])
21        return paths

ℹ

Complexity note: The time complexity is O(n) because we visit each node exactly once. The space complexity is O(n) due to the recursion stack and the storage of the paths.

1Understanding tree traversal is crucial for solving tree-related problems.
2Using lists for path construction can improve performance over string concatenation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.