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.

Solution:
Basic tree traverse question.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
        return traverse(root, sum);
    }
    public boolean traverse(TreeNode root, int remain){
        if(root.left == null && root.right == null){
            if(remain == root.val) return true;
            else return false;
        }
        Boolean leftResult = false;
        Boolean rightResult = false;
        if(root.left != null) leftResult = hasPathSum(root.left, remain - root.val);
        if(root.right != null) rightResult = hasPathSum(root.right, remain - root.val);
        return(leftResult || rightResult);
    }
}