How do you solve Path Sum on LeetCode?

Path Sum (LeetCode #112) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(h) space complexity. Key insight: The problem requires understanding of tree traversal techniques like DFS.

What is the brute force solution for Path Sum?

The brute force approach for Path Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves exploring all possible paths from the root to each leaf node and calculating their sums. If any path's sum matches the targetSum, we return true; otherwise, we return false. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Path Sum?

Path Sum uses the following concepts: Tree, Depth-First Search, Breadth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Recursion.

What are the interview tips for Path Sum?

Always clarify the definition of a leaf node with the interviewer. Discuss your thought process and approach before jumping into coding. Consider edge cases like empty trees or negative values.

What are common mistakes when solving Path Sum?

Not checking for empty trees at the start. Failing to correctly handle the case where the current node is a leaf.

#112

Path Sum

Easy

TreeDepth-First SearchBreadth-First SearchBinary TreeDepth-First SearchRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(h)

The optimal solution uses Depth-First Search (DFS) to traverse the tree while keeping track of the remaining targetSum. This approach efficiently checks each path without revisiting nodes.

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Algorithm

4 steps

1Step 1: Start DFS from the root node with the initial targetSum.
2Step 2: Subtract the current node's value from targetSum.
3Step 3: If a leaf node is reached, check if targetSum is zero.
4Step 4: Recursively call DFS for left and right children.

solution.py6 lines

1def hasPathSum(root, targetSum):
2    if not root:
3        return False
4    if not root.left and not root.right:
5        return targetSum == root.val
6    return hasPathSum(root.left, targetSum - root.val) or hasPathSum(root.right, targetSum - root.val)

ℹ

Complexity note: The time complexity is linear because we visit each node once. The space complexity is O(h) due to the recursive call stack, where h is the height of the tree.

1The problem requires understanding of tree traversal techniques like DFS.
2Identifying leaf nodes is crucial for determining valid paths.

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