112. Path Sum
Easy
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22
,
5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2
which sum is 22.
Accepted
485,008
Submissions
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/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public boolean hasPathSum(TreeNode root, int sum) { return helper(root, sum, 0); } boolean helper(TreeNode node, int sum, int cur){ if(node == null) return false; cur += node.val; if(node.left==null && node.right==null && cur==sum) return true; return(helper(node.left, sum, cur) || helper(node.right, sum, cur)); } }
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