112. Path Sum 🔗
Difficulty: Easy
- Tags: Binary Tree
, Depth-First Search (DFS)
, Recursion
Problem Statement 📜
Given the root of a binary tree and an integer targetSum
, return true
if the tree has a root-to-leaf path such that the sum of all the node values along the path equals targetSum
.
A leaf is a node with no children.
Examples 🌟
🔹 Example 1:

Input:
root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output:
true
Explanation:
The root-to-leaf path [5,4,11,2]
has a sum of 22
.
🔹 Example 2:

Input:
root = [1,2,3], targetSum = 5
Output:
false
Explanation:
There are two root-to-leaf paths:
Path
1 -> 2
: Sum is3
.Path
1 -> 3
: Sum is4
.
Neither equals 5
.
🔹 Example 3:
Input:
root = [], targetSum = 0
Output:
false
Explanation:
The tree is empty, so there are no root-to-leaf paths.
Constraints ⚙️
The number of nodes in the tree is in the range
[0, 5000]
.-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
Solution 💡
To determine if there is a root-to-leaf path with the given targetSum
:
Use Depth-First Search (DFS) to traverse the binary tree.
Subtract the current node's value from
targetSum
as you traverse.Check if:
The current node is a leaf.
The updated
targetSum
equals0
.
Java Solution
class Solution {
public boolean hasPathSum(TreeNode root, int targetSum) {
// Base case: if the tree is empty, there is no path
if (root == null) return false;
// If it's a leaf node, check if the targetSum matches the node's value
if (root.left == null && root.right == null) {
return targetSum == root.val;
}
// Recursively check left and right subtrees
int remainingSum = targetSum - root.val;
return hasPathSum(root.left, remainingSum) || hasPathSum(root.right, remainingSum);
}
}
Explanation of the Solution
Base Case:
If the root is
null
, the tree is empty, and no path exists.
Leaf Node Check:
If the node has no children, compare its value to the remaining
targetSum
.
Recursive Calls:
Subtract the current node's value from
targetSum
.Recursively check the left and right subtrees.
Combine Results:
Use the logical
OR
operator to combine results from both subtrees.
Time Complexity ⏳
O(n): Each node is visited once.
Space Complexity 💾
O(h): Where
h
is the height of the tree (due to recursion stack).
Follow-up Challenges 🧐
How would you modify this solution to return all paths with the target sum?
Can you solve this iteratively instead of using recursion?
You can find the full solution here.
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