Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.
A leaf is a node with no children.
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 with the target sum is shown.
Example 2:
Input: root = [1,2,3], targetSum = 5 Output: false Explanation: There two root-to-leaf paths in the tree: (1 --> 2): The sum is 3. (1 --> 3): The sum is 4. There is no root-to-leaf path with sum = 5.
Example 3:
Input: root = [], targetSum = 0 Output: false Explanation: Since the tree is empty, 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
What you need to find out is if there exists any path from root to leaf such that sum of those elements is equal to targetSum. As with any binary tree problem I started to try with recursion.
If there are no elements, then by default answer is false since there are no root to leaf paths.
Otherwise check if there is path in left subtree or right subtree whose sum is equal to targetSum minus value of current node. If either of these paths exists, then return true.
Below is the java code for the same.
public boolean hasPathSum(TreeNode root, int targetSum) {
if (root == null) return false;
else if (root.left == null && root.right == null) return root.val == targetSum;
else return hasPathSum(root.left, targetSum-root.val) || hasPathSum(root.right, targetSum-root.val);And below is the efficiency. It is 100% faster than any LeetCode submitted solution. 🙂
