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:
[0, 5000].-1000 <= Node.val <= 1000-1000 <= targetSum <= 1000
class Solution {
public boolean hasPathSum(TreeNode root, int targetSum) {
if(root == null){
return false ;
}
if(root.left == null && root.right == null && root.val == targetSum){
return true ;
}
boolean ans1 = hasPathSum(root.left , targetSum-root.val);
boolean ans2 = hasPathSum(root.right, targetSum-root.val);
return ans1 || ans2 ;
}
}
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