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.

 1 public class Solution {
 2 //    private static boolean flag=false;
 3 //    public static boolean hasPathSum(TreeNode root, int sum) {
 4 //    if(root==null){return flag;}
 5 ////    if(root.left==null&&root.right==null&&root.val==1&&sum==0)return false;
 6 //        F(0,root,sum);
 7 //    return flag;
 8 //    }
 9 //    public static void F(int cout,TreeNode node,int sum){
10 //    if(flag==true){return;}
11 //    cout+=node.val;
12 //    if(node.left==null&&node.right==null&&cout==sum){flag=true;return;}
13 //    if(node.left!=null){F(cout,node.left,sum);}
14 //    if(node.right!=null){F(cout,node.right,sum);}
15 //    }
16     public static boolean hasPathSum(TreeNode root, int sum) {
17     if (root == null)
18         return false;
19 
20     if (root.left == null && root.right == null)
21         return sum == root.val;
22 
23     return hasPathSum(root.left, sum - root.val)
24         || hasPathSum(root.right, sum - root.val);
25     }
26     public static class TreeNode {
27     int val;
28     TreeNode left;
29     TreeNode right;
30 
31     TreeNode(int x) {
32         val = x;
33     }
34     }
35     public static void main(String[]args){
36     TreeNode[] node=new TreeNode[5];
37     for(int i=0;i<node.length;i++){
38         node[i]=new TreeNode(i);
39     }
40     TreeNode node1=new TreeNode(1);
41     TreeNode node2=new TreeNode(2);
42 //    node1.left=node2;
43     node[0].left=node[1];
44     node[0].right=node[2];
45     node[1].left=node[3];
46     node[2].right=node[4];
47     System.out.println(hasPathSum(node1,2));
48     }
49 }