Path Sum

https://leetcode.com/problems/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.

public class Solution {
//    private static boolean flag=false;
//    public static boolean hasPathSum(TreeNode root, int sum) {
//    if(root==null){return flag;}
////    if(root.left==null&&root.right==null&&root.val==1&&sum==0)return false;
//        F(0,root,sum);
//    return flag;
//    }
//    public static void F(int cout,TreeNode node,int sum){
//    if(flag==true){return;}
//    cout+=node.val;
//    if(node.left==null&&node.right==null&&cout==sum){flag=true;return;}
//    if(node.left!=null){F(cout,node.left,sum);}
//    if(node.right!=null){F(cout,node.right,sum);}
//    }
    public static boolean hasPathSum(TreeNode root, int sum) {
    if (root == null)
        return false;

    if (root.left == null && root.right == null)
        return sum == root.val;

    return hasPathSum(root.left, sum - root.val)
        || hasPathSum(root.right, sum - root.val);
    }
    public static class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int x) {
        val = x;
    }
    }
    public static void main(String[]args){
    TreeNode[] node=new TreeNode[5];
    for(int i=0;i<node.length;i++){
        node[i]=new TreeNode(i);
    }
    TreeNode node1=new TreeNode(1);
    TreeNode node2=new TreeNode(2);
//    node1.left=node2;
    node[0].left=node[1];
    node[0].right=node[2];
    node[1].left=node[3];
    node[2].right=node[4];
    System.out.println(hasPathSum(node1,2));
    }
}