Path Sum

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.

For 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.

when reach the leaf node, check whether the sum equals to the target.

public boolean hasPathSum(TreeNode root, int sum) {
        if (root==null) {
            return false;
        }        
        return hasPathSum(root, 0, sum);
    }

    public boolean hasPathSum(TreeNode root, int currentSum, int sum) {
        if (root.left==null&&root.right==null) {
            if ((currentSum+root.val)==sum) {
                return true;
            } else {
                return false;
            }
        }

        boolean hasLeftPathSum = false;
        if (root.left!=null) {
            hasLeftPathSum = hasPathSum(root.left, currentSum+root.val, sum);
        }
        boolean hasRightPathSum = false;
        if (root.right!=null) {
            hasRightPathSum = hasPathSum(root.right, currentSum+root.val, sum);
        }
        return hasLeftPathSum||hasRightPathSum;
    }